Answer
$x\approx 1.0451$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition to our equation, we get:
$\log_{4}{77.2}=3x$
Next, recall the change of base formula (pg. 464):
$\log_{b}{m}=\dfrac{\log_{c}{m}}{\log_{c}{b}}$
Applying this formula to our last equation, we get:
$\dfrac{\log_{10}{77.2}}{\log_{10}{4}}=3x$
$\dfrac{1}{3}\cdot \dfrac{\log_{10}{77.2}}{\log_{10}{4}}=x$
$x\approx 1.0451$
We confirm that the answer works:
$4^{3\cdot1.0451}=77.2$
$4^{3.135}=77.2$
$77.2=77.2$
