Step(s) wrong in solving for $x$, $5^{x+2}=4^{1-x}$

Question

Kindly assist me in finding my error in solving for $x$:

$$5^{x+2}=4^{1-x}$$

My worksheet says that the answer cannot be expressed as an exact term, which is the same as saying the answer cannot be expressed in terms of $\log$. Additionally, my numerical value is incorrect.

Here are my steps:

$$5^x\cdot5^2=\frac{4^1}{4^x}$$

$$5^x4^x=\frac 4{25}$$

Taking $\log_4$ of both sides:

$$x\log_4 5+x=1-2\log_4 5$$ $$x(\log_4 5+1)=1-2\log_4 5$$ $$x=\frac{-2\log_4 5+1}{\log_4 5+1}$$

Performing long division:

$$x=-2-\frac 1{\log_45+1}=-2-(\log_45+1)^{-1}$$

From this, $-2-(\log_45+1)^{-1}$ is my exact solution.

$-2-(\log_45+1)^{-1}$ evaluates to around $-2.463$, but the solution says $x=-0.612$

What am I doing wrong? Thank you for the assistance!

Answer 1

After you perform long division

$$x = -2 + \frac{3}{\log_45+1}$$

Answer 2

You made a mistake here

$$x=\frac{-2\log_4 5+1}{\log_4 5+1}$$ $$x=\frac{-2\log_4 5-2+3}{\log_4 5+1}=\frac{-2(\log_4 5+1)+3}{\log_4 5+1}$$ $$x=-2 +\frac{3}{\log_4 5+1}$$