How do you solve $log(2x) - log(2^x) = log(x^2) - log(x)$ ?

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Answer 1 · 2016-01-09T01:34:49

$x=1$

Typically, it is difficult to solve equations that combine polynomials, logs, and exponentials.

To make progress with this equation, you must apply log rules.

$log(2x)-log(2^x)=log(x^2)-log(x)$

$log((2x)/(2^x))=log(x^2/x)$

Because you have two equal log statements, then the pieces inside must be equal as well...

$((2x)/(2^x))=(x^2/x)$

simplifying...

$(2x)/(2^x)=x$

doing algebra...

$(2x)/x=2^x$

$2=2^x$

$x=1$

How do you solve log(2x) - log(2^x) = log(x^2) - log(x)log(2x) - log(2^x) = log(x^2) - log(x)?