Question

Solve the equation. Help please!$x^{log_{3}x-4 } = 3^{-3} \\\\x^{log_{2}x } = 16$

Answer 1

First equation:

$x^{\log_3x-4}=3^{-3}$

$\log_3\left(x^{\log_3x-4}\right)=\log_33^{-3}$

$(\log_3x-4)\log_3x=-3\log_33$

$(\log_3x)^2-4\log_3x=-3$

$(\log_3x)^2-4\log_3x+3=0$

$(\log_3x-3)(\log_3x-1)=0$

$\log_3x-3=0\text{ or }\log_3x-1=0$

$\log_3x=3\text{ or }\log_3x=1$

$3^{\log_3x}=3^3\text{ or }3^{\log_3x}=3^1$

$x=27\text{ or }x=3$

Second equation:

$x^{\log_2x}=16$

$x^{\log_2x}=2^4$

$\log_2\left(x^{\log_2x}\right)=\log_22^4$

$\log_2x\log_2x=4\log_22$

$(\log_2x)^2=4$

$\log_2x=\pm2$

$2^{\log_2x}=2^{\pm2}$

$x=2^2\text{ or }x=2^{-2}$

$\boxed{x=4\text{ or }x=\dfrac14}$