Question

Hello, I need some help with this precalculus question for my homework, please HW Q6

Answer 1

`A)(87)/(8)`

Step-by-step explanation

let's remember some properites of the logarithms:

`\begin{gathered} \log_bM-\log_bN=\log_b((M)/(N)) \\ \log_aM=b\text{ , M= a}^b \end{gathered}`

so

Step 1

`\log_(2x+7)=1+\log_(x-8)`

a)

`\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((2x+7)/(x-8))=1 \end{gathered}`

b)now, convert the logarith into its exponential form( second property )

`\begin{gathered} log((2x+7)/(x-8))=1 \\ so\text{ } \\ (2x+7)/(x-8)=10^1 \\ (2x+7)/(x-8)=10 \end{gathered}`

c) finally, isolate x

`\begin{gathered} (2x+7)/(x-8)=10 \\ Multiply\text{ both sides by \lparen x-8\rparen} \\ (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} \\ (87)/(8)=(8x)/(8) \\ x=(87)/(8),\text{ x}\in\langle8,\infty\rangle \end{gathered}`

so, the answer is

`A)(87)/(8)`

I hope this helps you