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.

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

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

$4 = 41(x-4)$

$4 = 41x -164$

$4+164 = 41x$

$168 = 41x$

$x = 168/41$

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