#### Answer

$x\approx-17.531$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
5^{x+2}=2^{2x-1}
,$ take the logarithm of both sides. Use the laws of logarithms and the properties of equality to isolate the variable. Finally, Express the answer with $3$ decimal places.
$\bf{\text{Solution Details:}}$
Taking the natural logarithm of both sides, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\log5^{x+2}=\log2^{2x-1}
.\end{array}
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
(x+2)\log5=(2x-1)\log2
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(\log5)+2(\log5)=2x(\log2)-1(\log2)
\\\\
x\log5+2\log5=2x\log2-\log2
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
\log2+2\log5=2x\log2-x\log5
\\\\
\log2+2\log5=x(2\log2-\log5)
\\\\
\dfrac{\log2+2\log5}{2\log2-\log5}=x
\\\\
x=\dfrac{\log2+2\log5}{2\log2-\log5}
\\\\
x\approx-17.531
.\end{array}