## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\approx 6.0302\times 10^{17}$ Note: (If the RHS is 3.8 instead of 38, then $x\approx 4.7900$)
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 $x\approx 6.0302\times 10^{17}$