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

1 Answer
Jan 9, 2016

Answer:

#x=1#

Explanation:

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#