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

3 Answers
Sep 18, 2016

Answer:

#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#

Sep 18, 2016

Answer:

#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#

Sep 18, 2016

Answer:

#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)#.