## Intermediate Algebra (6th Edition)

$x=\dfrac{\log4+5\log5}{3\log5}\\\\ x\approx1.9538$
Getting the logarithm of both sides and then using the properties of logarithms, the value of $x$ in the expression $5^{3x-5}=4$ is \begin{array}{l} \log5^{3x-5}=\log4\\\\ (3x-5)\log5=\log4\\\\ 3x\log5-5\log5=\log4\\\\ 3x\log5=\log4+5\log5\\\\ x=\dfrac{\log4+5\log5}{3\log5}\\\\ x\approx1.9538 .\end{array}