(x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

(2x - 1)³ = -64

=> (2x - 1)³ = -4³

=> 2x - 1 = -4

=> 2x = -4 + 1

=> 2x = -3

=> x = -3/2

( x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

64 . 4^x = 16⁸

<=> 4³. 4^x = 4¹⁶

=> 3 + x = 16

<=> x = 13

Vậy x = 13

2^x.16² = 1024

<=> 2^x. 2⁸ = 2¹⁰

=> x + 8 = 10

<=> x = 2

Vậy x = 2

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I am doing homework

a, (\(\frac{1}{2}\))^x = \(\frac{1}{64}\)

<=> (\(\frac{1}{2}\))^x = (\(\frac{1}{2}\))⁶

<=> x = 6

b, 3⁵ :3^x = 9

<=> 3⁵ :3^x = 3² 

<=> 3^x = 3⁵ :3²

<=> 3^x = 3³

<=> x = 3

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a,\(\left(\frac{1}{2}\right)^x=\frac{1}{64}\)

\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^6\)

\(\Rightarrow x=6\)

b,\(3^5:3^x=9\)

\(3^5:3^x=3^2\)

\(3^x=3^5:3^2\)

\(3^x=3^3\)

\(\Rightarrow x=3\)

1. ( x-2)² = 81

( x-2)² = 9²

=>x-2=9

x=9+2

x=11

\(\left(x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+6\\x=-10\end{matrix}\right.\)

\(x+2=8\)

\(x=6\)

\(\left(x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)