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