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

##### 3 Answers

#### 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.

#### 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#

#### Explanation:

We have:

Using the laws of logarithms:

Therefore, the solution to the equation is