Answer
$x=\dfrac{\log15}{\log9}\approx1.232487$
Work Step by Step
$3^{2x-1}=5$
Using the product of powers rule, we can write the left side of the equation like this:
$(3^{2x})(3^{-1})=5$
We know that $(a^{b})^{c}=a^{bc}$ and $a^{-n}=\dfrac{1}{a^{n}}$. Applying this to the left side, it becomes:
$(3^{2})^{x}(\dfrac{1}{3})=5$
$(9^{x})(\dfrac{1}{3})=5$
Take the fraction's denominator to multiply the right side of the equation:
$9^{x}=15$
Apply $\log$ to both sides:
$\log9^{x}=\log5$
The exponent $x$ can be taken down to multiply in front of the $\log$:
$x\log9=\log15$
Solve for $x$:
$x=\dfrac{\log15}{\log9}\approx1.232487$