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log(subscript 3)(5x-1)=log(subscript 3)(x+7)
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\[\log_{3}(5x-1)=\log_{3}(x+7) \]
since they have the same bases just solve it as a normal equation 5x-1 = x+7 4x = 8 x=2
$$\bf \large { log_{3}(5x-1)=log_{3}(x+7)\\ \text{using the cancellation rule of}\\ a^{log_ax} = x\\ \color{blue}{3}^{log_{\color{blue}{3}}(5x-1)}=\color{blue}{3}^{log_{\color{blue}{3}}(x+7)} \implies 5x-1 = x+7 } $$
then as cmolina19 said, just solve for "x" :)
You"re great teacher. I'm a new fan. Thank you for the help.
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