College Algebra (10th Edition)

$x = \dfrac{\log{(\frac{3}{8})} + 7}{2}$
Divide both sides of the equation by 8 to obtain: $\dfrac{8 \cdot 10^{2x-7}}{8} = \dfrac{3}{8} \\10^{2x-7}=\dfrac{3}{8}$ Take the common logarithm of both sides to obtain: $\log{(10^{2x-7})}=\log{(\frac{3}{8})}$ Note that $\log{(10^x}) = x$. Thus, the equation above is equivalent to: $2x-7=\log{(\frac{3}{8})}$ Add $7$ to both sides of the equation to obtain: $2x-7+7 = \log{(\frac{3}{8})} + 7 \\2x = \log{(\frac{3}{8})} + 7$ Divide by 2 on both sides of the equation to obtain: $\dfrac{2x}{2} = \dfrac{\log{(\frac{3}{8})} + 7}{2} \\x = \dfrac{\log{(\frac{3}{8})} + 7}{2}$