How do you solve #log_7x-log_7(x-1)=1#?

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Answer 1 · 2016-07-18T10:25:47

#x = 7/6#

When working with logs, the terms must all be in either log form or as numbers, but not a combination of both.

note: #log_10 10 = 1 and log_2 2 = 1 and log_7 7 =1 # etc.#

In order to have all the terms written as logs with the same base, the given equation can be changed to

#log_7x-log_7(x-1)=log_7 7#

If log terms are added, they can be written as the log of one number:

#log_7 (x/(x-1)) = log_7 7#

But if log A = log B, then A = B

So we now have: #x/(x-1) =7# and can treat it as a normal linear equation:

#x = 7(x-1)#
#x = 7x -7#
#7 = 6x#
#x = 7/6#