How can i solve (lnx)^2 = 1 ? I know that there is a way to move the logarithm to the other side of the equalty and it ends up like e^something, but i can't figure out the way to do that.

Question

lgbasallote · Answer

first..take the square root of both sides $$\implies \sqrt{(\ln x)^2} = \sqrt 1$$
simplify...
$$\implies \ln x = \pm 1$$
change to log form
$$\implies \log_e x = \pm 1$$
change to exponential form
$$e^{\pm 1} = x$$
therefore$$x = e ; \; x = \frac 1e$$
does that help?

phi · Answer

e^(ln x)=x

anonymous · Answer

yeah, thank you lgbasallote that help me a lot :) I was actually stuck in the last step.

lgbasallote · Answer

you have been helped in the name of Jesus.

lgbasallote · Answer

^i always wanted to say that

lgbasallote · Answer

and welcome