#### Answer

$x\approx-2.487$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
6^{x-3}=3^{4x+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 logarithm of both sides, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\log6^{x-3}=\log3^{4x+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-3)\log6=(4x+1)\log3
.\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(\log6)-3(\log6)=4x(\log3)+1(\log3)
\\\\
x\log6-3\log6=4x\log3+\log3
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
-\log3-3\log6=4x\log3-x\log6
\\\\
-\log3-3\log6=x(4\log3-\log6)
\\\\
\dfrac{-\log3-3\log6}{4\log3-\log6}=x
\\\\
x=\dfrac{-\log3-3\log6}{4\log3-\log6}
\\\\
x\approx-2.487
.\end{array}