How do you solve $\log _ { 12} ( 2x + 13) = \log _ { 12} ( 4x - 7)$ ?

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Answer 1 · 2018-04-21T03:18:41

Take antilog of $12$ on both sides we get :-

$2x+13=4x-7$

$rArr13+7=4x-2x$

$rArr20=2x$

$:.x=10$

Also $(2x+13)$ and $(4x-7)$ are Positive for $x=20$ , thus this value of $x$ is allowed.

How do you solve \log _ { 12} ( 2x + 13) = \log _ { 12} ( 4x - 7)\log _ { 12} ( 2x + 13) = \log _ { 12} ( 4x - 7)?