Explanation
let's remember some properites of the logarithms:
[tex]\begin{gathered} \log_bM-\log_bN=\log_b(\frac{M}{N}) \\ \log_aM=b\text{ , M= a}^b \end{gathered}[/tex]so
Step 1
[tex]\log_(2x+7)=1+\log_(x-8)[/tex]a)
[tex]\begin{gathered} \log_(2x+7)=1+\log_(x-8) \\ subtract\text{ }\log_(x+8)\text{ in both sides} \\ log(2x+7)-log(x-8)=1+log(X-8)-log(x-8) \\ \begin{equation*} log(2x+7)-log(x-8)=1 \end{equation*} \\ apply\text{ the propery} \\ log(\frac{2x+7}{x-8})=1 \end{gathered}[/tex]b)now, convert the logarith into its exponential form( second property )
[tex]\begin{gathered} log(\frac{2x+7}{x-8})=1 \\ so\text{ } \\ \frac{2x+7}{x-8}=10^1 \\ \frac{2x+7}{x-8}=10 \end{gathered}[/tex]c) finally, isolate x
[tex]\begin{gathered} \frac{2x+7}{x-8}=10 \\ Multiply\text{ both sides by \lparen x-8\rparen} \\ \frac{2x+7}{x-8}*(x-8)=10(x-8) \\ 2x+7=10x-80 \\ subtract\text{ 2x in both sides} \\ 2x+7-2x=10x-80-2x \\ 7=8x-80 \\ add\text{ 80 in both sides} \\ 7+80=8x-80+80 \\ 87=8x \\ divide\text{ both sides by 8} \\ \frac{87}{8}=\frac{8x}{8} \\ x=\frac{87}{8},\text{ x}\in\langle8,\infty\rangle \end{gathered}[/tex]so, the answer is
[tex]A)\frac{87}{8}[/tex]I hope this helps you
