Master the properties of exponentiation and radicals in ℝ with clear definitions and special cases. Learn about natural, fractional, and negative exponents, sign rules for powers, exponential equations, and simplifying radicals. Perfect for math students!
a n = { a if n = 1 a ⋅ a ⋅ … ⋅ a ⏟ n times if n ∈ N , n ≥ 2 \boxed{a^n =
\begin{cases}
a & \text{if } n = 1 \\
\underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ times}} & \text{if } n \in \mathbb{N}, n \geq 2
\end{cases}}
a n = ⎩ ⎨ ⎧ a n times a ⋅ a ⋅ … ⋅ a if n = 1 if n ∈ N , n ≥ 2
The expression:
( x y ) ⋅ ( x y ) ⋅ ( x y ) ⋯ ( x y ) ⏟ ( 5 + 3 ) times ≠ ( x y ) 5 + 3 {\underbrace{(\sqrt{xy})\cdot(\sqrt{xy})\cdot(\sqrt{xy}) \cdots (\sqrt{xy})}_{(\sqrt{5}+\sqrt{3}) \text{ times}}\neq (\sqrt{xy})^{\sqrt{5}+\sqrt{3}}}
( 5 + 3 ) times ( x y ) ⋅ ( x y ) ⋅ ( x y ) ⋯ ( x y ) = ( x y ) 5 + 3
is not defined , since ( 5 + 3 ) {(\sqrt{5}+\sqrt{3})} ( 5 + 3 ) not a natural number.
Sign rule for powers with negative bases:
( − b ) 2 n = + b 2 n ; ∀ n ∈ Z + (-b)^{2n} = +b^{2n}; \quad \forall n \in \mathbb{Z}^+
( − b ) 2 n = + b 2 n ; ∀ n ∈ Z +
( − b ) 2 n + 1 = − b 2 n + 1 ; ∀ n ∈ Z + (-b)^{2n+1} = -b^{2n+1}; \quad \forall n \in \mathbb{Z}^+
( − b ) 2 n + 1 = − b 2 n + 1 ; ∀ n ∈ Z +
3 4 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⏟ 4 times = 81 3^4 = \underbrace{3 \cdot 3 \cdot 3 \cdot 3}_{4 \text{ times}} = 81 3 4 = 4 times 3 ⋅ 3 ⋅ 3 ⋅ 3 = 81 ( 1 2 ) 3 = 1 2 ⋅ 1 2 ⋅ 1 2 ⏟ 3 times = 1 8 \left(\dfrac{1}{2}\right)^3 = \underbrace{\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}}_{3 \text{ times}} = \dfrac{1}{8} ( 2 1 ) 3 = 3 times 2 1 ⋅ 2 1 ⋅ 2 1 = 8 1 ( − 2 ) 5 = ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) ⏟ 5 times = − 32 (-2)^5 = \underbrace{(-2) \cdot (-2) \cdot (-2) \cdot (-2) \cdot (-2)}_{5 \text{ times}} = -32 ( − 2 ) 5 = 5 times ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) ⋅ ( − 2 ) = − 32 x 3 = x ⋅ x ⋅ x ⏟ 3 times x^3 = \underbrace{x \cdot x \cdot x}_{3 \text{ times}} x 3 = 3 times x ⋅ x ⋅ x 7 1 = 7 (by definition, if n = 1 ) 7^1 = 7 \quad \text{(by definition, if } n = 1\text{)} 7 1 = 7 (by definition, if n = 1 ) ( − 1 ) 6 = ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⏟ 6 times = 1 (-1)^6 = \underbrace{(-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1)}_{6 \text{ times}} = 1 ( − 1 ) 6 = 6 times ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) ⋅ ( − 1 ) = 1
a 0 = 1 ∀ a ∈ R ∖ { 0 } \boxed{
a^0 = 1 \quad \forall a \in \mathbb{R} \setminus \{0\}}
a 0 = 1 ∀ a ∈ R ∖ { 0 }
0 0 0^0 0 0 indeterminate form .
5 0 = 1 5^0 = 1 5 0 = 1 ( − 7 ) 0 = 1 (-7)^0 = 1 ( − 7 ) 0 = 1 ( 3.14 ) 0 = 1 (3.14)^0 = 1 ( 3.14 ) 0 = 1 ( 2 3 ) 0 = 1 \left(\frac{2}{3}\right)^0 = 1 ( 3 2 ) 0 = 1 ( x 2 + 1 ) 0 = 1 for any x ∈ R (x^2 + 1)^0 = 1 \quad \text{for any } x \in \mathbb{R} ( x 2 + 1 ) 0 = 1 for any x ∈ R ( a ⋅ b ⋅ c ) 0 = 1 if a , b , c ≠ 0 (a \cdot b \cdot c)^0 = 1 \quad \text{if } a, b, c \neq 0 ( a ⋅ b ⋅ c ) 0 = 1 if a , b , c = 0
a − n = 1 a n ∀ a ∈ R ∖ { 0 } , n ∈ N \boxed{a^{-n} = \frac{1}{a^n} \quad \forall a \in \mathbb{R} \setminus \{0\}, n \in \mathbb{N}}
a − n = a n 1 ∀ a ∈ R ∖ { 0 } , n ∈ N
0 n 0^n 0 n is undefined for n ∈ N n \in \mathbb{N} n ∈ N
2 − 3 = 1 2 3 = 1 8 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8} 2 − 3 = 2 3 1 = 8 1 5 − 2 = 1 5 2 = 1 25 5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25} 5 − 2 = 5 2 1 = 25 1 ( − 4 ) − 3 = 1 ( − 4 ) 3 = 1 − 64 = − 1 64 (-4)^{-3} = \dfrac{1}{(-4)^3} = \dfrac{1}{-64} = -\dfrac{1}{64} ( − 4 ) − 3 = ( − 4 ) 3 1 = − 64 1 = − 64 1 ( 1 3 ) − 2 = 1 ( 1 3 ) 2 = 1 1 9 = 9 \left(\dfrac{1}{3}\right)^{-2} = \dfrac{1}{\left(\dfrac{1}{3}\right)^2} = \dfrac{1}{\dfrac{1}{9}} = 9 ( 3 1 ) − 2 = ( 3 1 ) 2 1 = 9 1 1 = 9 x − 4 = 1 x 4 if x ≠ 0 x^{-4} = \dfrac{1}{x^4} \quad \text{if } x \neq 0 x − 4 = x 4 1 if x = 0 ( 2 a ) − 1 = 1 2 a if a ≠ 0 (2a)^{-1} = \dfrac{1}{2a} \quad \text{if } a \neq 0 ( 2 a ) − 1 = 2 a 1 if a = 0
a m / n = a m n = ( a n ) m ∀ n ∈ N , n ≥ 2 \boxed{
a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m \quad \forall n \in \mathbb{N}, n \geq 2}
a m / n = n a m = ( n a ) m ∀ n ∈ N , n ≥ 2
( + ) 2 n + 1 = ( + ) \sqrt[2n+1]{(+)} = (+)
2 n + 1 ( + ) = ( + )
( − ) 2 n + 1 = ( − ) \sqrt[2n+1]{(-)} = (-)
2 n + 1 ( − ) = ( − )
( + ) 2 n = ( + ) \sqrt[2n]{(+)} = (+)
2 n ( + ) = ( + )
( − ) 2 n = undefined in R \sqrt[2n]{(-)} = \text{undefined in }\mathbb{R}
2 n ( − ) = undefined in R
8 2 / 3 = 8 2 3 = 64 3 = 4 8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4 8 2/3 = 3 8 2 = 3 64 = 4 16 3 / 4 = ( 16 4 ) 3 = 2 3 = 8 16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8 1 6 3/4 = ( 4 16 ) 3 = 2 3 = 8 27 1 / 3 = 27 3 = 3 27^{1/3} = \sqrt[3]{27} = 3 2 7 1/3 = 3 27 = 3 4 3 / 2 = ( 4 ) 3 = 2 3 = 8 (here 2 = ) 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \quad \text{(here } \sqrt[2]{\ } = \sqrt{\ }\text{)} 4 3/2 = ( 4 ) 3 = 2 3 = 8 (here 2 = ) x 5 / 2 = x 5 = ( x ) 5 if x ≥ 0 x^{5/2} = \sqrt{x^5} = (\sqrt{x})^5 \quad \text{if } x \geq 0 x 5/2 = x 5 = ( x ) 5 if x ≥ 0 ( 1 9 ) 1 / 2 = 1 9 = 1 3 \left(\dfrac{1}{9}\right)^{1/2} = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3} ( 9 1 ) 1/2 = 9 1 = 3 1
P = a n ; a ∈ R , n ∈ N , P ∈ R \boxed{P = a^n; \quad a \in \mathbb{R}, n \in \mathbb{N}, P \in \mathbb{R}}
P = a n ; a ∈ R , n ∈ N , P ∈ R
Where:
a a a n n n P P P
a m ⋅ a n = a m + n a ∈ R , m , n ∈ N \boxed{a^m \cdot a^n = a^{m+n} \quad a \in \mathbb{R}, m, n \in \mathbb{N}}
a m ⋅ a n = a m + n a ∈ R , m , n ∈ N
2 3 ⋅ 2 4 = 2 3 + 4 = 2 7 = 128 2^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128 2 3 ⋅ 2 4 = 2 3 + 4 = 2 7 = 128 x 5 ⋅ x 2 = x 5 + 2 = x 7 x^5 \cdot x^2 = x^{5+2} = x^7 x 5 ⋅ x 2 = x 5 + 2 = x 7 ( − 3 ) 2 ⋅ ( − 3 ) 6 = ( − 3 ) 2 + 6 = ( − 3 ) 8 = 6561 (-3)^2 \cdot (-3)^6 = (-3)^{2+6} = (-3)^8 = 6561 ( − 3 ) 2 ⋅ ( − 3 ) 6 = ( − 3 ) 2 + 6 = ( − 3 ) 8 = 6561 a 4 ⋅ a 1 = a 4 + 1 = a 5 a^4 \cdot a^1 = a^{4+1} = a^5 a 4 ⋅ a 1 = a 4 + 1 = a 5 ( 1 2 ) 3 ⋅ ( 1 2 ) 5 = ( 1 2 ) 3 + 5 = ( 1 2 ) 8 = 1 256 \left(\dfrac{1}{2}\right)^3 \cdot \left(\dfrac{1}{2}\right)^5 = \left(\dfrac{1}{2}\right)^{3+5} = \left(\dfrac{1}{2}\right)^8 = \dfrac{1}{256} ( 2 1 ) 3 ⋅ ( 2 1 ) 5 = ( 2 1 ) 3 + 5 = ( 2 1 ) 8 = 256 1 y n ⋅ y 7 = y n + 7 y^{n} \cdot y^{7} = y^{n+7} y n ⋅ y 7 = y n + 7
( a m ) n = a m ⋅ n a ∈ R , m , n ∈ N \boxed{(a^m)^n = a^{m \cdot n} \quad a \in \mathbb{R}, m, n \in \mathbb{N}}
( a m ) n = a m ⋅ n a ∈ R , m , n ∈ N
( ( ( a m ) n ) r ) s = a m ⋅ n ⋅ r ⋅ s \boxed{\left(((a^m)^n)^r\right)^s = a^{m \cdot n \cdot r \cdot s}}
( (( a m ) n ) r ) s = a m ⋅ n ⋅ r ⋅ s
( 2 3 ) 4 = 2 3 ⋅ 4 = 2 12 = 4096 (2^3)^4 = 2^{3 \cdot 4} = 2^{12} = 4096 ( 2 3 ) 4 = 2 3 ⋅ 4 = 2 12 = 4096 ( x 2 ) 5 = x 2 ⋅ 5 = x 10 (x^2)^5 = x^{2 \cdot 5} = x^{10} ( x 2 ) 5 = x 2 ⋅ 5 = x 10 ( ( − 3 ) 4 ) 2 = ( − 3 ) 4 ⋅ 2 = ( − 3 ) 8 = 6561 ((-3)^4)^2 = (-3)^{4 \cdot 2} = (-3)^8 = 6561 (( − 3 ) 4 ) 2 = ( − 3 ) 4 ⋅ 2 = ( − 3 ) 8 = 6561 ( a 3 ) 1 = a 3 ⋅ 1 = a 3 (a^3)^1 = a^{3 \cdot 1} = a^3 ( a 3 ) 1 = a 3 ⋅ 1 = a 3 ( ( 1 2 ) 2 ) 3 = ( 1 2 ) 2 ⋅ 3 = ( 1 2 ) 6 = 1 64 \left(\left(\dfrac{1}{2}\right)^2\right)^3= \left(\dfrac{1}{2}\right)^{2 \cdot 3} = \left(\dfrac{1}{2}\right)^6 = \dfrac{1}{64} ( ( 2 1 ) 2 ) 3 = ( 2 1 ) 2 ⋅ 3 = ( 2 1 ) 6 = 64 1 ( y m ) 7 = y m ⋅ 7 = y 7 m (y^m)^7 = y^{m \cdot 7} = y^{7m} ( y m ) 7 = y m ⋅ 7 = y 7 m
( a ⋅ b ) n = a n ⋅ b n a , b ∈ R , n ∈ N \boxed{(a \cdot b)^n = a^n \cdot b^n \quad a, b \in \mathbb{R}, n \in \mathbb{N}}
( a ⋅ b ) n = a n ⋅ b n a , b ∈ R , n ∈ N
( x a ⋅ y b ) n = x a ⋅ n ⋅ y b ⋅ n \boxed{(x^a \cdot y^b)^n = x^{a \cdot n} \cdot y^{b \cdot n}}
( x a ⋅ y b ) n = x a ⋅ n ⋅ y b ⋅ n
( 3 ⋅ 4 ) 2 = 3 2 ⋅ 4 2 = 9 ⋅ 16 = 144 (3 \cdot 4)^2 = 3^2 \cdot 4^2 = 9 \cdot 16 = 144 ( 3 ⋅ 4 ) 2 = 3 2 ⋅ 4 2 = 9 ⋅ 16 = 144 ( 2 x ) 3 = 2 3 ⋅ x 3 = 8 x 3 (2x)^3 = 2^3 \cdot x^3 = 8x^3 ( 2 x ) 3 = 2 3 ⋅ x 3 = 8 x 3 ( − 5 ⋅ 2 ) 4 = ( − 5 ) 4 ⋅ 2 4 = 625 ⋅ 16 = 10000 (-5 \cdot 2)^4 = (-5)^4 \cdot 2^4 = 625 \cdot 16 = 10000 ( − 5 ⋅ 2 ) 4 = ( − 5 ) 4 ⋅ 2 4 = 625 ⋅ 16 = 10000 ( a b ) 5 = a 5 ⋅ b 5 (ab)^5 = a^5 \cdot b^5 ( ab ) 5 = a 5 ⋅ b 5 ( 1 2 ⋅ y ) 3 = ( 1 2 ) 3 ⋅ y 3 = 1 8 y 3 \left(\dfrac{1}{2} \cdot y\right)^3 = \left(\dfrac{1}{2}\right)^3 \cdot y^3 = \dfrac{1}{8}y^3 ( 2 1 ⋅ y ) 3 = ( 2 1 ) 3 ⋅ y 3 = 8 1 y 3 ( x y 2 ) 4 = x 4 ⋅ ( y 2 ) 4 = x 4 ⋅ y 8 (xy^2)^4 = x^4 \cdot (y^2)^4 = x^4 \cdot y^8 ( x y 2 ) 4 = x 4 ⋅ ( y 2 ) 4 = x 4 ⋅ y 8
a m a n = a m − n m , n ∈ N , m ≥ n , a ∈ R ∖ { 0 } \boxed{\frac{a^m}{a^n} = a^{m-n} \quad m, n \in \mathbb{N}, m \geq n, a \in \mathbb{R} \setminus \{0\}}
a n a m = a m − n m , n ∈ N , m ≥ n , a ∈ R ∖ { 0 }
5 7 5 3 = 5 7 − 3 = 5 4 = 625 \dfrac{5^7}{5^3} = 5^{7-3} = 5^4 = 625 5 3 5 7 = 5 7 − 3 = 5 4 = 625 x 8 x 5 = x 8 − 5 = x 3 \dfrac{x^8}{x^5} = x^{8-5} = x^3 x 5 x 8 = x 8 − 5 = x 3 ( − 2 ) 6 ( − 2 ) 2 = ( − 2 ) 6 − 2 = ( − 2 ) 4 = 16 \dfrac{(-2)^6}{(-2)^2} = (-2)^{6-2} = (-2)^4 = 16 ( − 2 ) 2 ( − 2 ) 6 = ( − 2 ) 6 − 2 = ( − 2 ) 4 = 16 a 10 a 10 = a 10 − 10 = a 0 = 1 (if a ≠ 0 ) \dfrac{a^{10}}{a^{10}} = a^{10-10} = a^0 = 1 \quad \text{(if } a \neq 0\text{)} a 10 a 10 = a 10 − 10 = a 0 = 1 (if a = 0 ) 3 5 3 2 = 3 5 − 2 = 3 3 = 27 \dfrac{3^5}{3^2} = 3^{5-2} = 3^3 = 27 3 2 3 5 = 3 5 − 2 = 3 3 = 27 y n + 4 y n = y ( n + 4 ) − n = y 4 \dfrac{y^{n+4}}{y^n} = y^{(n+4)-n} = y^4 y n y n + 4 = y ( n + 4 ) − n = y 4
( a b ) n = a n b n n ∈ N , b ∈ R ∖ { 0 } \boxed{\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad n \in \mathbb{N}, b \in \mathbb{R} \setminus \{0\}}
( b a ) n = b n a n n ∈ N , b ∈ R ∖ { 0 }
( 2 3 ) 4 = 2 4 3 4 = 16 81 \left(\dfrac{2}{3}\right)^4 = \dfrac{2^4}{3^4} = \dfrac{16}{81} ( 3 2 ) 4 = 3 4 2 4 = 81 16 ( 5 2 ) 3 = 5 3 2 3 = 125 8 \left(\dfrac{5}{2}\right)^3 = \dfrac{5^3}{2^3} = \dfrac{125}{8} ( 2 5 ) 3 = 2 3 5 3 = 8 125 ( x 4 ) 2 = x 2 4 2 = x 2 16 if x ∈ R \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{4^2} = \dfrac{x^2}{16} \quad \text{if } x \in \mathbb{R} ( 4 x ) 2 = 4 2 x 2 = 16 x 2 if x ∈ R ( − 3 5 ) 3 = ( − 3 ) 3 5 3 = − 27 125 \left(\dfrac{-3}{5}\right)^3 = \dfrac{(-3)^3}{5^3} = \dfrac{-27}{125} ( 5 − 3 ) 3 = 5 3 ( − 3 ) 3 = 125 − 27 ( a b ) 5 = a 5 b 5 if b ≠ 0 \left(\dfrac{a}{b}\right)^5 = \dfrac{a^5}{b^5} \quad \text{if } b \neq 0 ( b a ) 5 = b 5 a 5 if b = 0 ( 1 x ) n = 1 n x n = 1 x n if x ≠ 0 \left(\dfrac{1}{x}\right)^n = \dfrac{1^n}{x^n} = \dfrac{1}{x^n} \quad \text{if } x \neq 0 ( x 1 ) n = x n 1 n = x n 1 if x = 0
( a b ) − n = ( b a ) n = b n a n a , b ≠ 0 \boxed{\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n = \frac{b^n}{a^n} \quad a, b \neq 0}
( b a ) − n = ( a b ) n = a n b n a , b = 0
( 2 3 ) − 4 = ( 3 2 ) 4 = 3 4 2 4 = 81 16 \left(\dfrac{2}{3}\right)^{-4} = \left(\dfrac{3}{2}\right)^4 = \dfrac{3^4}{2^4} = \dfrac{81}{16} ( 3 2 ) − 4 = ( 2 3 ) 4 = 2 4 3 4 = 16 81 ( 5 x ) − 2 = ( x 5 ) 2 = x 2 25 if x ≠ 0 \left(\dfrac{5}{x}\right)^{-2} = \left(\dfrac{x}{5}\right)^2 = \dfrac{x^2}{25} \quad \text{if } x \neq 0 ( x 5 ) − 2 = ( 5 x ) 2 = 25 x 2 if x = 0 ( 1 4 ) − 3 = ( 4 1 ) 3 = 4 3 = 64 \left(\dfrac{1}{4}\right)^{-3} = \left(\dfrac{4}{1}\right)^3 = 4^3 = 64 ( 4 1 ) − 3 = ( 1 4 ) 3 = 4 3 = 64 ( a b ) − 1 = b a if a , b ≠ 0 \left(\dfrac{a}{b}\right)^{-1} = \dfrac{b}{a} \quad \text{if } a, b \neq 0 ( b a ) − 1 = a b if a , b = 0 ( − 2 7 ) − 3 = ( 7 − 2 ) 3 = 7 3 ( − 2 ) 3 = 343 − 8 = − 343 8 \left(\dfrac{-2}{7}\right)^{-3} = \left(\dfrac{7}{-2}\right)^3 = \dfrac{7^3}{(-2)^3} = \dfrac{343}{-8} = -\dfrac{343}{8} ( 7 − 2 ) − 3 = ( − 2 7 ) 3 = ( − 2 ) 3 7 3 = − 8 343 = − 8 343 ( x y ) − 5 = y 5 x 5 if x , y ≠ 0 \left(\dfrac{x}{y}\right)^{-5} = \dfrac{y^5}{x^5} \quad \text{if } x, y \neq 0 ( y x ) − 5 = x 5 y 5 if x , y = 0
x a b c = x a m = x n = z \boxed{x^{a^{b^c}} = x^{a^m}= x^n= z}
x a b c = x a m = x n = z
( x m ) n ≠ x m n {(x^m)^n \neq x^{m^n}}
( x m ) n = x m n
( x + y ) n ≠ x n + y n {(x+y)^n \neq x^n+y^n}
( x + y ) n = x n + y n
2 2 3 1 = 2 2 3 = 2 8 = 256 2^{2^{3^1}} = 2^{2^3} = 2^8 = 256 2 2 3 1 = 2 2 3 = 2 8 = 256 3 2 2 2 = 3 2 4 = 3 16 = 43 , 046 , 721 3^{2^{2^2}} = 3^{2^4} = 3^{16} = 43,\!046,\!721 3 2 2 2 = 3 2 4 = 3 16 = 43 , 046 , 721 5 1 4 2 = 5 1 16 = 5 1 = 5 5^{1^{4^2}} = 5^{1^{16}} = 5^1 = 5 5 1 4 2 = 5 1 16 = 5 1 = 5 10 3 1 5 = 10 3 1 = 10 3 = 1 , 000 10^{3^{1^5}} = 10^{3^1} = 10^3 = 1,\!000 1 0 3 1 5 = 1 0 3 1 = 1 0 3 = 1 , 000 2 3 2 1 = 2 3 2 = 2 9 = 512 2^{3^{2^1}} = 2^{3^2} = 2^9 = 512 2 3 2 1 = 2 3 2 = 2 9 = 512 4 2 3 0 = 4 2 1 = 4 2 = 16 4^{2^{3^0}} = 4^{2^1} = 4^2 = 16 4 2 3 0 = 4 2 1 = 4 2 = 16
x 2 k 2 k = ∣ x ∣ = { x if: x > 0 0 if: x = 0 − x if: x < 0 \sqrt[2k]{x^{2k}} = |x| =
\begin{cases}
x & \text{if: } x > 0 \\
0 & \text{if: } x = 0 \\
-x & \text{if: } x < 0
\end{cases}
2 k x 2 k = ∣ x ∣ = ⎩ ⎨ ⎧ x 0 − x if: x > 0 if: x = 0 if: x < 0
Most commonly used powers
2 2 = 4 2^2 = 4 2 2 = 4 3 2 = 9 3^2 = 9 3 2 = 9 4 2 = 16 4^2 = 16 4 2 = 16 8 2 = 64 8^2 = 64 8 2 = 64
2 3 = 8 2^3 = 8 2 3 = 8 3 3 = 27 3^3 = 27 3 3 = 27 4 3 = 64 4^3 = 64 4 3 = 64 8 3 = 512 8^3 = 512 8 3 = 512
2 4 = 16 2^4 = 16 2 4 = 16 3 4 = 81 3^4 = 81 3 4 = 81 5 2 = 25 5^2 = 25 5 2 = 25 9 2 = 81 9^2 = 81 9 2 = 81
2 5 = 32 2^5 = 32 2 5 = 32 3 5 = 243 3^5 = 243 3 5 = 243 5 3 = 125 5^3 = 125 5 3 = 125 9 3 = 729 9^3 = 729 9 3 = 729
2 6 = 64 2^6 = 64 2 6 = 64 3 6 = 729 3^6 = 729 3 6 = 729 5 4 = 625 5^4 = 625 5 4 = 625 10 2 = 100 10^2 = 100 1 0 2 = 100
2 7 = 128 2^7 = 128 2 7 = 128 5 5 = 3125 5^5 = 3125 5 5 = 3125 10 3 = 1000 10^3 = 1000 1 0 3 = 1000
2 8 = 256 2^8 = 256 2 8 = 256 6 2 = 36 6^2 = 36 6 2 = 36 11 2 = 121 11^2 = 121 1 1 2 = 121
2 9 = 512 2^9 = 512 2 9 = 512 6 3 = 216 6^3 = 216 6 3 = 216 11 3 = 1331 11^3 = 1331 1 1 3 = 1331
2 10 = 1024 2^{10} = 1024 2 10 = 1024 6 4 = 1296 6^4 = 1296 6 4 = 1296 12 2 = 144 12^2 = 144 1 2 2 = 144
7 2 = 49 7^2 = 49 7 2 = 49 12 3 = 1728 12^3 = 1728 1 2 3 = 1728
7 3 = 343 7^3 = 343 7 3 = 343
7 4 = 2401 7^4 = 2401 7 4 = 2401
y = x n ⟺ y n = x \boxed{y = \sqrt[n]{x} \iff y^n = x }
y = n x ⟺ y n = x
Where:
x x x x ∈ R , x ≥ 0 x \in \mathbb{R}, x \geq 0 x ∈ R , x ≥ 0 n n n n n n n ∈ N , n ≥ 2 n \in \mathbb{N}, n \geq 2 n ∈ N , n ≥ 2 y y y y ≥ 0 y \geq 0 y ≥ 0 n n n
a ⋅ b n = a n ⋅ b n \boxed{\sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b}}
n a ⋅ b = n a ⋅ n b
If n n n a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0
4 ⋅ 9 = 4 ⋅ 9 = 2 ⋅ 3 = 6 \sqrt{4 \cdot 9} = \sqrt{4} \cdot \sqrt{9} = 2 \cdot 3 = 6 4 ⋅ 9 = 4 ⋅ 9 = 2 ⋅ 3 = 6 8 ⋅ 27 3 = 8 3 ⋅ 27 3 = 2 ⋅ 3 = 6 \sqrt[3]{8 \cdot 27} = \sqrt[3]{8} \cdot \sqrt[3]{27} = 2 \cdot 3 = 6 3 8 ⋅ 27 = 3 8 ⋅ 3 27 = 2 ⋅ 3 = 6 16 ⋅ 81 4 = 16 4 ⋅ 81 4 = 2 ⋅ 3 = 6 \sqrt[4]{16 \cdot 81} = \sqrt[4]{16} \cdot \sqrt[4]{81} = 2 \cdot 3 = 6 4 16 ⋅ 81 = 4 16 ⋅ 4 81 = 2 ⋅ 3 = 6 x ⋅ y = x ⋅ y if x ≥ 0 , y ≥ 0 \sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y} \quad \text{if } x \geq 0,\ y \geq 0 x ⋅ y = x ⋅ y if x ≥ 0 , y ≥ 0 a 5 ⋅ b 10 5 = a 5 5 ⋅ b 10 5 = a ⋅ b 2 \sqrt[5]{a^5 \cdot b^{10}} = \sqrt[5]{a^5} \cdot \sqrt[5]{b^{10}} = a \cdot b^2 5 a 5 ⋅ b 10 = 5 a 5 ⋅ 5 b 10 = a ⋅ b 2 − 8 ⋅ 64 3 = − 8 3 ⋅ 64 3 = ( − 2 ) ⋅ 4 = − 8 \sqrt[3]{-8 \cdot 64} = \sqrt[3]{-8} \cdot \sqrt[3]{64} = (-2) \cdot 4 = -8 3 − 8 ⋅ 64 = 3 − 8 ⋅ 3 64 = ( − 2 ) ⋅ 4 = − 8
a b n = a n b n b ≠ 0 \boxed{\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \quad b \neq 0}
n b a = n b n a b = 0
If n n n a ≥ 0 a \geq 0 a ≥ 0 b > 0 b > 0 b > 0
9 16 = 9 16 = 3 4 \sqrt{\dfrac{9}{16}} = \dfrac{\sqrt{9}}{\sqrt{16}} = \dfrac{3}{4} 16 9 = 16 9 = 4 3 27 125 3 = 27 3 125 3 = 3 5 \sqrt[3]{\dfrac{27}{125}} = \dfrac{\sqrt[3]{27}}{\sqrt[3]{125}} = \dfrac{3}{5} 3 125 27 = 3 125 3 27 = 5 3 1 81 4 = 1 4 81 4 = 1 3 \sqrt[4]{\dfrac{1}{81}} = \dfrac{\sqrt[4]{1}}{\sqrt[4]{81}} = \dfrac{1}{3} 4 81 1 = 4 81 4 1 = 3 1 x 5 32 5 = x 5 5 32 5 = x 2 if x ∈ R \sqrt[5]{\dfrac{x^5}{32}} = \dfrac{\sqrt[5]{x^5}}{\sqrt[5]{32}} = \dfrac{x}{2} \quad \text{if } x \in \mathbb{R} 5 32 x 5 = 5 32 5 x 5 = 2 x if x ∈ R a b = a b if a ≥ 0 , b > 0 \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} \quad \text{if } a \geq 0,\ b > 0 b a = b a if a ≥ 0 , b > 0 − 8 27 3 = − 8 3 27 3 = − 2 3 \sqrt[3]{\dfrac{-8}{27}} = \dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}} = \dfrac{-2}{3} 3 27 − 8 = 3 27 3 − 8 = 3 − 2
a n m = a m ⋅ n m , n ∈ N \boxed{\sqrt[m]{\sqrt[n]{a}} = \sqrt[m \cdot n]{a} \quad m, n \in \mathbb{N}}
m n a = m ⋅ n a m , n ∈ N
If m ⋅ n m \cdot n m ⋅ n a ≥ 0 a \geq 0 a ≥ 0
a r s n m = a m ⋅ n ⋅ s ⋅ r \boxed{\sqrt[m]{\sqrt[n]{\sqrt[s]{\sqrt[r]{a}}}} = \sqrt[m \cdot n \cdot s \cdot r]{a}}
m n s r a = m ⋅ n ⋅ s ⋅ r a
8 3 2 = 8 6 = 2 3 6 = 2 3 / 6 = 2 1 / 2 = 2 \sqrt[2]{\sqrt[3]{8}} = \sqrt[6]{8} = \sqrt[6]{2^3} = 2^{3/6} = 2^{1/2} = \sqrt{2} 2 3 8 = 6 8 = 6 2 3 = 2 3/6 = 2 1/2 = 2 x 4 3 = x 12 for x ≥ 0 \sqrt[3]{\sqrt[4]{x}} = \sqrt[12]{x} \quad \text{for } x \geq 0 3 4 x = 12 x for x ≥ 0 16 = 16 2 2 = 16 4 = 2 \sqrt{\sqrt{16}} = \sqrt[2]{\sqrt[2]{16}} = \sqrt[4]{16} = 2 16 = 2 2 16 = 4 16 = 2 a 5 = a 10 if a ≥ 0 \sqrt[5]{\sqrt{a}} = \sqrt[10]{a} \quad \text{if } a \geq 0 5 a = 10 a if a ≥ 0 64 3 4 = 64 12 = 2 6 12 = 2 6 / 12 = 2 1 / 2 = 2 \sqrt[4]{\sqrt[3]{64}} = \sqrt[12]{64} = \sqrt[12]{2^6} = 2^{6/12} = 2^{1/2} = \sqrt{2} 4 3 64 = 12 64 = 12 2 6 = 2 6/12 = 2 1/2 = 2 x = x 2 2 2 = x 8 for x ≥ 0 \sqrt{\sqrt{\sqrt{x}}} = \sqrt[2]{\sqrt[2]{\sqrt[2]{x}}} = \sqrt[8]{x} \quad \text{for } x \geq 0 x = 2 2 2 x = 8 x for x ≥ 0
( a n ) m = a m n \boxed{\left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}}
( n a ) m = n a m
x a c a b = x c b If a b is even, then x ∈ R 0 + \boxed{\sqrt[ab]{x^{ac}} = \sqrt[b]{x^c} \quad \text{If } ab \text{ is even, then } x \in \mathbb{R}_0^+}
ab x a c = b x c If ab is even, then x ∈ R 0 +
( 2 3 ) 6 = 2 6 3 = 64 3 = 4 \left(\sqrt[3]{2}\right)^6 = \sqrt[3]{2^6} = \sqrt[3]{64} = 4 ( 3 2 ) 6 = 3 2 6 = 3 64 = 4 ( 5 ) 4 = 5 4 = 625 = 25 \left(\sqrt{5}\right)^4 = \sqrt{5^4} = \sqrt{625} = 25 ( 5 ) 4 = 5 4 = 625 = 25 ( 3 4 ) 2 = 3 2 4 = 9 4 \left(\sqrt[4]{3}\right)^2 = \sqrt[4]{3^2} = \sqrt[4]{9} ( 4 3 ) 2 = 4 3 2 = 4 9 ( x n ) 3 = x 3 n if x ≥ 0 when n is even \left(\sqrt[n]{x}\right)^3 = \sqrt[n]{x^3} \quad \text{if } x \geq 0 \text{ when } n \text{ is even} ( n x ) 3 = n x 3 if x ≥ 0 when n is even ( − 32 5 ) 3 = ( − 32 ) 3 5 = − 32768 5 = − 8 \left(\sqrt[5]{-32}\right)^3 = \sqrt[5]{(-32)^3} = \sqrt[5]{-32768} = -8 ( 5 − 32 ) 3 = 5 ( − 32 ) 3 = 5 − 32768 = − 8 ( a 2 + b 2 ) 2 = ( a 2 + b 2 ) 2 = ∣ a 2 + b 2 ∣ = a 2 + b 2 \left(\sqrt{a^2 + b^2}\right)^2 = \sqrt{(a^2 + b^2)^2} = |a^2 + b^2| = a^2 + b^2 ( a 2 + b 2 ) 2 = ( a 2 + b 2 ) 2 = ∣ a 2 + b 2 ∣ = a 2 + b 2
x a ⋅ x b ⋅ x c p m n = x ( a m + b ) p + c n m p \sqrt[n]{x^{a}\cdot \sqrt[m]{x^{b}\cdot \sqrt[p]{x^c}}} = \sqrt[nmp]{x^{(am + b)p+c}}
n x a ⋅ m x b ⋅ p x c = nm p x ( am + b ) p + c
x a ÷ x b ÷ x c p m n = x ( a m − b ) p + c n m p \sqrt[n]{x^{a}\div \sqrt[m]{x^{b}\div \sqrt[p]{x^c}}} = \sqrt[nmp]{x^{(am - b)p+c}}
n x a ÷ m x b ÷ p x c = nm p x ( am − b ) p + c
x ⋅ y ⋅ z p n m = x m ⋅ y m ⋅ n ⋅ z n ⋅ m ⋅ p \sqrt[m]{x \cdot \sqrt[n]{y \cdot \sqrt[p]{z}}} = \sqrt[m]{x} \cdot \sqrt[m \cdot n]{y} \cdot \sqrt[n \cdot m \cdot p]{z}
m x ⋅ n y ⋅ p z = m x ⋅ m ⋅ n y ⋅ n ⋅ m ⋅ p z
x a ⋅ y b ⋅ z c p n m = x a m ⋅ y b m ⋅ n ⋅ z c n ⋅ m ⋅ p \sqrt[m]{x^a \cdot \sqrt[n]{y^b \cdot \sqrt[p]{z^c}}} = \sqrt[m]{x^a} \cdot \sqrt[m \cdot n]{y^b} \cdot \sqrt[n \cdot m \cdot p]{z^c}
m x a ⋅ n y b ⋅ p z c = m x a ⋅ m ⋅ n y b ⋅ n ⋅ m ⋅ p z c
x a ÷ y b ÷ z c ÷ w d ÷ v e ÷ u f s r q p n m = a x m ⋅ c z m n p ⋅ e v m n p q r b y m n ⋅ d w m n p q ⋅ f u m n p q r s \sqrt[m]{x^{a}\div \sqrt[n]{y^{b}\div \sqrt[p]{z^{c}\div \sqrt[q]{w^{d}\div \sqrt[r]{v^{e}\div \sqrt[s]{u^f}}}}}} = \frac{\cfrac{a}{x^m} \cdot \cfrac{c}{z^{mnp}} \cdot \cfrac{e}{v^{mnpqr}}}{\cfrac{b}{y^{mn}} \cdot \cfrac{d}{w^{mnpq}} \cdot \cfrac{f}{u^{mnpqrs}}}
m x a ÷ n y b ÷ p z c ÷ q w d ÷ r v e ÷ s u f = y mn b ⋅ w mn pq d ⋅ u mn pq rs f x m a ⋅ z mn p c ⋅ v mn pq r e
Radical Type
Simplified Form
Radical Form
Square Roots 2 2 2\sqrt{2} 2 2 8 \sqrt{8} 8
3 2 3\sqrt{2} 3 2 18 \sqrt{18} 18
4 2 4\sqrt{2} 4 2 32 \sqrt{32} 32
5 2 5\sqrt{2} 5 2 50 \sqrt{50} 50
3 3 3\sqrt{3} 3 3 27 \sqrt{27} 27
4 3 4\sqrt{3} 4 3 48 \sqrt{48} 48
5 3 5\sqrt{3} 5 3 75 \sqrt{75} 75
3 5 3\sqrt{5} 3 5 45 \sqrt{45} 45
4 5 4\sqrt{5} 4 5 80 \sqrt{80} 80
2 5 2\sqrt{5} 2 5 20 \sqrt{20} 20
Cube Roots 2 2 3 2\sqrt[3]{2} 2 3 2 16 3 \sqrt[3]{16} 3 16
3 2 3 3\sqrt[3]{2} 3 3 2 54 3 \sqrt[3]{54} 3 54
2 3 3 2\sqrt[3]{3} 2 3 3 24 3 \sqrt[3]{24} 3 24
3 3 3 3\sqrt[3]{3} 3 3 3 81 3 \sqrt[3]{81} 3 81
Approximate Values 2 \sqrt{2} 2 1.4142... 1.4142... 1.4142...
3 \sqrt{3} 3 1.732... 1.732... 1.732...
5 \sqrt{5} 5 2.236... 2.236... 2.236...
x m ⋅ y p n = x m n ⋅ y p n x^m \cdot \sqrt[n]{y^p} = \sqrt[n]{x^{mn} \cdot y^p}
x m ⋅ n y p = n x mn ⋅ y p
x m ⋅ x n = x n + m m n \sqrt[m]{x} \cdot \sqrt[n]{x} = \sqrt[mn]{x^{n+m}}
m x ⋅ n x = mn x n + m
x m x n = x n − m m n , x ≠ 0 \frac{\sqrt[m]{x}}{\sqrt[n]{x}} = \sqrt[mn]{x^{n-m}}, \, x \neq 0
n x m x = mn x n − m , x = 0
1 z c y b x a = 1 x ⋅ 1 y ⋅ 1 z c b a \sqrt[a]{\frac{\sqrt[b]{\cfrac{\sqrt[c]{\cfrac{1}{z}}}{y}}}{x}} = \sqrt[a]{\frac{1}{x}\cdot \sqrt[b]{\frac{1}{y}\cdot \sqrt[c]{\frac{1}{z}}}}
a x b y c z 1 = a x 1 ⋅ b y 1 ⋅ c z 1
a ⋅ b ⋅ c p n m d ⋅ e ⋅ f p n m = a d ⋅ b e ⋅ c f p n m \frac{\sqrt[m]{a\cdot \sqrt[n]{b\cdot \sqrt[p]{c}} } }{\sqrt[m]{d\cdot \sqrt[n]{e\cdot \sqrt[p]{f}} } } = \sqrt[m]{\frac{a}{d}\cdot \sqrt[n]{\frac{b}{e}\cdot \sqrt[p]{\frac{c}{f}}}}
m d ⋅ n e ⋅ p f m a ⋅ n b ⋅ p c = m d a ⋅ n e b ⋅ p f c
x y z p n m = x m y n z p \sqrt[m]{x^{ \sqrt[n]{y^{ \sqrt[p]{z}}}}} = \sqrt[m]{x}^{\sqrt[n]{y}^{\sqrt[p]{z}}}
m x n y p z = m x n y p z
Expressions with a Fixed Number of Radicals
… x n n n n ⏟ m radicals = x n m \underbrace{\sqrt[n]{ \sqrt[n]{ \sqrt[n]{ \dots \sqrt[n]{x}}}}}_{\text{m radicals}} = \sqrt[n^m]{x}
m radicals n n n … n x = n m x
x ⋅ x ⋅ x ⋅ . . . ⋅ x n n n n ⏟ m radicals = x n m − 1 n − 1 n m \underbrace{\sqrt[n]{x \cdot \sqrt[n]{x \cdot \sqrt[n]{x \cdot ...\cdot \sqrt[n]{x}}}}}_{\text{m radicals}} = \sqrt[n^m]{x^{\cfrac{n^m - 1}{n-1}}}
m radicals n x ⋅ n x ⋅ n x ⋅ ... ⋅ n x = n m x n − 1 n m − 1
x ÷ x ÷ x ÷ . . . ÷ x n n n n ⏟ m radicals = { x n m − 1 n + 1 n m if m is even , x n m + 1 n + 1 n m if m is odd . \underbrace{\sqrt[n]{x\div \sqrt[n]{x\div \sqrt[n]{x\div ... \div \sqrt[n]{x}}}}}_{\text{m radicals}} =
\begin{cases}
\sqrt[n^m]{x^{\cfrac{n^m - 1}{n + 1}}} & \text{if m is even}, \\
\sqrt[n^m]{x^{\cfrac{n^m + 1}{n + 1}}} & \text{if m is odd}.
\end{cases}
m radicals n x ÷ n x ÷ n x ÷ ... ÷ n x = ⎩ ⎨ ⎧ n m x n + 1 n m − 1 n m x n + 1 n m + 1 if m is even , if m is odd .
Expressions with Infinite Radicals
x ⋅ ( x + 1 ) + x ⋅ ( x + 1 ) + x ⋅ ( x + 1 ) + … = x + 1 \sqrt{x\cdot(x+1)+ \sqrt{x\cdot(x+1)+ \sqrt{x\cdot(x+1)+ \dots } }} = x+1
x ⋅ ( x + 1 ) + x ⋅ ( x + 1 ) + x ⋅ ( x + 1 ) + … = x + 1
x ⋅ ( x + 1 ) − x ⋅ ( x + 1 ) − x ⋅ ( x + 1 ) − … = x \sqrt{x\cdot(x+1)- \sqrt{x\cdot(x+1)- \sqrt{x\cdot(x+1)- \dots } }} = x
x ⋅ ( x + 1 ) − x ⋅ ( x + 1 ) − x ⋅ ( x + 1 ) − … = x
x ⋅ x ⋅ x ⋅ . . . n n n = x n − 1 \sqrt[n]{x\cdot \sqrt[n]{x\cdot \sqrt[n]{x\cdot...} }} = \sqrt[n-1]{x}
n x ⋅ n x ⋅ n x ⋅ ... = n − 1 x
x ÷ x ÷ x ÷ . . . n n n = x n + 1 \sqrt[n]{x\div \sqrt[n]{x\div \sqrt[n]{x\div...} }} = \sqrt[n+1]{x}
n x ÷ n x ÷ n x ÷ ... = n + 1 x
a ⋅ a ⋅ a ⋅ a ⋅ … m n m n = a m + 1 m ⋅ n − 1 \quad \sqrt[n]{a \cdot \sqrt[m]{a \cdot \sqrt[n]{a \cdot \sqrt[m]{a\cdot \dots} }}} = \sqrt[m \cdot n - 1]{a^{m+1}}
n a ⋅ m a ⋅ n a ⋅ m a ⋅ … = m ⋅ n − 1 a m + 1
a ⋅ b ⋅ a ⋅ b ⋅ … m n m n = a m ⋅ b m ⋅ n − 1 \quad \sqrt[n]{a \cdot \sqrt[m]{b \cdot \sqrt[n]{a \cdot \sqrt[m]{b\cdot \dots} }}} = \sqrt[m \cdot n - 1]{a^m \cdot b}
n a ⋅ m b ⋅ n a ⋅ m b ⋅ … = m ⋅ n − 1 a m ⋅ b
a + a + a + … = 1 + 4 a + 1 2 \quad \sqrt{a + \sqrt{a + \sqrt{a + \dots}}} = \frac{1 + \sqrt{4a + 1}}{2}
a + a + a + … = 2 1 + 4 a + 1
a − a − a − … = − 1 + 4 a + 1 2 \quad \sqrt{a - \sqrt{a - \sqrt{a - \dots}}} = \frac{-1 + \sqrt{4a + 1}}{2}
a − a − a − … = 2 − 1 + 4 a + 1
Infinite Power Towers
x x x . . . x n = n → x = n n ; x ≠ 0 x^{x^{x^{{.}^{{.}^{{.}^{x^n}}}}}} = n \rightarrow x = \sqrt[n]{n}; \, x \neq 0
x x x . . . x n = n → x = n n ; x = 0
b a b a b a b a . . . = c → b = c \sqrt[a]{b}^{\sqrt[a]{b}^{\sqrt[a]{b}^{\sqrt[a]{b}^{.^{.^{.}}}}}} = c \rightarrow b = c
a b a b a b a b . . . = c → b = c
a m = a n ⟹ m = n ; a > 0 ∧ a ≠ 1 a^m = a^n \implies m = n; \quad a > 0 \land a \neq 1
a m = a n ⟹ m = n ; a > 0 ∧ a = 1
2 x = 2 5 ⟹ x = 5 2^{x} = 2^{5} \implies x = 5 2 x = 2 5 ⟹ x = 5 3 2 y = 3 8 ⟹ 2 y = 8 ⟹ y = 4 3^{2y} = 3^{8} \implies 2y = 8 \implies y = 4 3 2 y = 3 8 ⟹ 2 y = 8 ⟹ y = 4 10 3 z − 1 = 10 z + 7 ⟹ 3 z − 1 = z + 7 ⟹ z = 4 10^{3z - 1} = 10^{z + 7} \implies 3z - 1 = z + 7 \implies z = 4 1 0 3 z − 1 = 1 0 z + 7 ⟹ 3 z − 1 = z + 7 ⟹ z = 4 ( 1 2 ) t = ( 1 2 ) 4 ⟹ t = 4 \left(\frac{1}{2}\right)^{t} = \left(\frac{1}{2}\right)^{4} \implies t = 4 ( 2 1 ) t = ( 2 1 ) 4 ⟹ t = 4 7 a + 2 = 7 2 a − 3 ⟹ a + 2 = 2 a − 3 ⟹ a = 5 7^{a+2} = 7^{2a - 3} \implies a + 2 = 2a - 3 \implies a = 5 7 a + 2 = 7 2 a − 3 ⟹ a + 2 = 2 a − 3 ⟹ a = 5 e 3 x = e x + 10 ⟹ 3 x = x + 10 ⟹ x = 5 e^{3x} = e^{x + 10} \implies 3x = x + 10 \implies x = 5 e 3 x = e x + 10 ⟹ 3 x = x + 10 ⟹ x = 5
a m = b m ⟹ a = b ; m ≠ 0 , a > 0 ∧ b > 0 a^m = b^m \implies a = b; \quad m \neq 0, a > 0 \land b > 0
a m = b m ⟹ a = b ; m = 0 , a > 0 ∧ b > 0
x 3 = 5 3 ⟹ x = 5 x^3 = 5^3 \implies x = 5 x 3 = 5 3 ⟹ x = 5 ( 2 y ) 4 = 6 4 ⟹ 2 y = 6 ⟹ y = 3 (2y)^4 = 6^4 \implies 2y = 6 \implies y = 3 ( 2 y ) 4 = 6 4 ⟹ 2 y = 6 ⟹ y = 3 a 7 = b 7 ⟹ a = b a^7 = b^7 \implies a = b a 7 = b 7 ⟹ a = b ( x + 1 ) 2 = ( 3 x − 1 ) 2 (x+1)^2 = (3x - 1)^2 ( x + 1 ) 2 = ( 3 x − 1 ) 2 x + 1 > 0 x+1 > 0 x + 1 > 0 3 x − 1 > 0 ⟹ x + 1 = 3 x − 1 ⟹ x = 1 3x - 1 > 0 \implies x+1 = 3x - 1 \implies x = 1 3 x − 1 > 0 ⟹ x + 1 = 3 x − 1 ⟹ x = 1 ( 4 z ) 5 = ( 2 z + 6 ) 5 ⟹ 4 z = 2 z + 6 ⟹ z = 3 (4z)^5 = (2z + 6)^5 \implies 4z = 2z + 6 \implies z = 3 ( 4 z ) 5 = ( 2 z + 6 ) 5 ⟹ 4 z = 2 z + 6 ⟹ z = 3 ( a 2 ) 6 = ( 3 2 ) 6 ⟹ a 2 = 3 2 ⟹ a = 3 \left(\frac{a}{2}\right)^6 = \left(\frac{3}{2}\right)^6 \implies \frac{a}{2} = \frac{3}{2} \implies a = 3 ( 2 a ) 6 = ( 2 3 ) 6 ⟹ 2 a = 2 3 ⟹ a = 3
a m = b n ⟹ m = n = 0 a^m = b^n \implies m = n = 0
a m = b n ⟹ m = n = 0
2 x = 7 y 2^x = 7^y 2 x = 7 y 2 x = 1 ⟹ x = 0 2^x = 1 \implies x = 0 2 x = 1 ⟹ x = 0 7 y = 1 ⟹ y = 0 7^y = 1 \implies y = 0 7 y = 1 ⟹ y = 0 3 a + 1 = 5 b − 2 3^{a+1} = 5^{b-2} 3 a + 1 = 5 b − 2 a + 1 = 0 a+1 = 0 a + 1 = 0 b − 2 = 0 ⟹ a = − 1 b-2 = 0 \implies a = -1 b − 2 = 0 ⟹ a = − 1 b = 2 b = 2 b = 2 10 m = 9 n = 1 ⟹ m = 0 10^m = 9^n = 1 \implies m = 0 1 0 m = 9 n = 1 ⟹ m = 0 n = 0 n = 0 n = 0 ( 1 3 ) 2 x = 4 y + 1 = 1 ⟹ 2 x = 0 \left(\frac{1}{3}\right)^{2x} = 4^{y+1} = 1 \implies 2x = 0 ( 3 1 ) 2 x = 4 y + 1 = 1 ⟹ 2 x = 0 y + 1 = 0 ⟹ x = 0 y+1 = 0 \implies x = 0 y + 1 = 0 ⟹ x = 0 y = − 1 y = -1 y = − 1 π t = e s = 1 ⟹ t = 0 \pi^{t} = e^{s} = 1 \implies t = 0 π t = e s = 1 ⟹ t = 0 s = 0 s = 0 s = 0 6 x − 3 = 8 2 − y = 1 ⟹ x − 3 = 0 6^{x-3} = 8^{2-y} = 1 \implies x - 3 = 0 6 x − 3 = 8 2 − y = 1 ⟹ x − 3 = 0 2 − y = 0 ⟹ x = 3 2 - y = 0 \implies x = 3 2 − y = 0 ⟹ x = 3 y = 2 y = 2 y = 2
a a = b b ⟹ a = b ; a ≠ 0 ∧ b ≠ 0 a^a = b^b \implies a = b; \quad a \neq 0 \land b \neq 0
a a = b b ⟹ a = b ; a = 0 ∧ b = 0
x x = 2 2 ⟹ x = 2 x^x = 2^2 \implies x = 2 x x = 2 2 ⟹ x = 2 y y = 3 3 ⟹ y = 3 y^y = 3^3 \implies y = 3 y y = 3 3 ⟹ y = 3 z z = 1 1 ⟹ z = 1 z^z = 1^1 \implies z = 1 z z = 1 1 ⟹ z = 1 a a = ( 1 2 ) 1 / 2 ⟹ a = 1 2 a^a = \left(\frac{1}{2}\right)^{1/2} \implies a = \frac{1}{2} a a = ( 2 1 ) 1/2 ⟹ a = 2 1 b b = 4 4 ⟹ b = 4 b^b = 4^4 \implies b = 4 b b = 4 4 ⟹ b = 4 t t = 5 5 ⟹ t = 5 t^t = 5^5 \implies t = 5 t t = 5 5 ⟹ t = 5
a a a + m = a a a + n ⟹ m = n a^{a^{a+m}} = a^{a^{a+n}} \implies m = n
a a a + m = a a a + n ⟹ m = n
2 2 2 + m = 2 2 2 + 3 ⟹ m = 3 2^{2^{2+m}} = 2^{2^{2+3}} \implies m = 3 2 2 2 + m = 2 2 2 + 3 ⟹ m = 3 3 3 3 + x = 3 3 3 + 1 ⟹ x = 1 3^{3^{3+x}} = 3^{3^{3+1}} \implies x = 1 3 3 3 + x = 3 3 3 + 1 ⟹ x = 1 2 2 2 + y = 2 2 2 + y + 0 ⟹ y = y 2^{2^{2+y}} = 2^{2^{2+y+0}} \implies y = y 2 2 2 + y = 2 2 2 + y + 0 ⟹ y = y m = n m = n m = n 4 4 4 + a = 4 4 4 + 0 ⟹ a = 0 4^{4^{4+a}} = 4^{4^{4+0}} \implies a = 0 4 4 4 + a = 4 4 4 + 0 ⟹ a = 0 5 5 5 + t = 5 5 5 + 2 ⟹ t = 2 5^{5^{5+t}} = 5^{5^{5+2}} \implies t = 2 5 5 5 + t = 5 5 5 + 2 ⟹ t = 2 10 10 10 + z = 10 10 10 + 7 ⟹ z = 7 10^{10^{10+z}} = 10^{10^{10+7}} \implies z = 7 1 0 1 0 10 + z = 1 0 1 0 10 + 7 ⟹ z = 7
x x m = m ⟹ x = m m x^{x^m} = m \implies x = \sqrt[m]{m}
x x m = m ⟹ x = m m
Exponential Form :
x x = a n ⟹ x = a x^x = a^n \implies x = a
x x = a n ⟹ x = a
Radical Form :
x x = n ⟹ x = n x x^x = n \implies x = \sqrt[x]{n}
x x = n ⟹ x = x n
