Question

Please solve this with work.

Answer 1

`\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log(6+7x)-\log(3x-2)=\log(5)\implies \log\left( \cfrac{6+7x}{3x-2} \right)=\log(5) \\\\\\ \cfrac{6+7x}{3x-2}=5\implies 6+7x=15x-10\implies 6=8x-10 \\\\\\ 16=8x\implies \cfrac{16}{8}=x\implies 2=x`

Answer 2

x = 2

Explanation:

The relevant rule of logarithms is ...

log(a/b) = log(a) -log(b)

This lets us combine the logs on the left to get ...

log((6 +7x)/(3x -2)) = log(5)

Taking anti-logs gives ...

(6 +7x)/(3x -2) = 5

6 +7x = 5(3x -2) . . . . . multiply by (3x -2)

7x +6 = 15x -10 . . . . . eliminate parentheses

8x = 16 . . . . . . . . . . . add 10-7x to both sides

x = 2 . . . . . . . . . . . . . divide by 8

The solution is x = 2.

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The graph shows the left side of the equation is equal to the right side for x=2.