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

Question

7347 views around the world

You can reuse this answer
Creative Commons License

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#