Question

How do you solve `log_6 5-log_6(x-7)=1`?

Answer 1

`x = 47/6`

Explanation:

Two principles at work here...

If you are subtracting the logs, divide the numbers.

All numbers or all logs, not both! Change 1 to a log.

`log_6 5-log_6(x-7)=1`

`log_6 (5/(x-7)) = log_6 6 " "larr` if log A = log B, then A=B

`5/(x-7) = 6`

`5 = 6(x-7)`

`5 = 6x -42`

`47 = 6x`

`47/6 = x`

Answer 2

`x=47/6`

Explanation:

Using the `color(blue)"laws of logarithms"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(logx-logy=log(x/y))color(white)(a/a)|)))`

`rArrlog_6 5-log_6(x-7)=log_6(5/(x-7))`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(log_b a=nhArra=b^n)color(white)(a/a)|)))`

`rArrlog_6(5/(x-7))=1rArr5/(x-7)=6^1=6`

`rArr6(x-7)=5rArr6x-42=5`

`rArr6x=47rArrx=47/6`

Answer 3

`x = (47) / (6)`

Explanation:

We have: `log_(6)(5) - log_(6)(x - 7) = 1`

Using the laws of logarithms:

`=> log_(6)((5) / (x - 7)) = 1`

`=> (5) / (x - 7) = 6^(1)`

`=> (5) / (x - 7) = 6`

`=> 5 = 6 (x - 7)`

`=> x - 7 = (5) / (6)`

`=> x = (47) / (6)`

Therefore, the solution to the equation is `x = (47) / (6)`.