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

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Answer 1 · 2016-01-09T01:34:49

$x = 1$ $x=1$

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

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

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

$log (\frac{2 x}{2^{x}}) = log (\frac{x^{2}}{x})$ $log((2x)/(2^x))=log(x^2/x)$

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

$(\frac{2 x}{2^{x}}) = (\frac{x^{2}}{x})$ $((2x)/(2^x))=(x^2/x)$

simplifying...

$\frac{2 x}{2^{x}} = x$ $(2x)/(2^x)=x$

doing algebra...

$\frac{2 x}{x} = 2^{x}$ $(2x)/x=2^x$

$2 = 2^{x}$ $2=2^x$

$x = 1$ $x=1$

How do you solve log(2x)−log(2x)=log(x2)−log(x)log(2x) - log(2^x) = log(x^2) - log(x)?