How do you solve #log_7 4-log_7(x-4)=log_7 41#?

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Answer 1 · 2016-08-07T18:27:38

#x = 168/41#

#log_7 4-log_7(x-4)=log_7 41#

If log terms are being subtracted, the numbers are being divided.

#log_7(4/(x-4))= log_7 41#

#:. 4/(x-4)= 41 " if " logA = logB, "then " A = B#

#4 = 41(x-4)#

#4 = 41x -164#

#4+164 = 41x#

#168 = 41x#

#x = 168/41#