Answer
$x\approx 6.0302\times 10^{17}$
Note:
(If the RHS is 3.8 instead of 38, then $x\approx 4.7900$)
Work Step by Step
Apply $\log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log 275+\log x^{2}=38\qquad$ ... subtract $\log 275$
$\log x^{2}=38-\log 275\qquad$ ... apply $\log_{a}M^{p}=p\cdot\log_{a}M$
$ 2\log x=38-\log 275\qquad$ ... divide with 2,
$\displaystyle \log x=\frac{38-\log 275}{2}\qquad$ ... apply inverse function $10^{(...)}$
$x=10^{(\frac{38-\log 275}{2})}$
Before we calculate, I feel that the RHS of the problem should be 3.8 instead of 38, as written in the text.
So, I offer two answers:
If the RHS = 38, $x\approx 6.0302268916\times 10^{17}$
If the RHS = 3.8, $x\approx 4.78997948176$
The text has 38, so $x\approx 6.0302\times 10^{17}$
