Answer
$x\approx 0.8505$
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:
$\log_{9}{42}=2x$
Next, recall the change of base formula (pg. 464):
$\log_{b}{m}=\dfrac{\log_{c}{m}}{\log_{c}{b}}$
Applying this property, we get:
$\dfrac{\log_{10}{42}}{\log_{10}{9}}=2x$
$x=\dfrac{1}{2} \cdot \dfrac{\log_{10}{42}}{\log_{10}{9}}$
$x\approx 0.8505$
We confirm that the answer works:
$9^{2\cdot 0.8505}=42$
$9^{1.701}=42$
$42=42$
