How to solve for #x e^(ln2x) = 12#?

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Answer 1 · 2018-03-08T12:15:00

See below.

I am not sure if this is supposed to be:

#xe^(ln(2x))=12# #bbor# #e^(ln(2x))=12#

I will do this for both, and then you can choose which is correct.

For:

#xe^(ln(2x))=12#

The law of logarithms state that:

#b^(log_ba)=a#

#:.#

#x(2x)=12#

#2x^2=12#

#x^2=12/2=6#

#x=+-sqrt(6)#

Only #sqrt(6)# is valid for real numbers:

#ln(x)# requires #x>0#

So:

#color(blue)(x=sqrt(6))#

For:

#e^(ln(2x))=12#

#2x=12#

#x=12/2=6#

#color(blue)(x=6)#