Logarithmic equations.

[axrw]

I am so sorry. You guys must just be sick of me by now.

I have two equations I can't figure out how to solve.

\(\displaystyle \sqrt{\ln x} = \ln \sqrt{x}\)

I started off squaring both sides:

\(\displaystyle \ln x = \frac{1}{4}(\ln x)(\ln x)\)

Then dividing by 1/4 and one of the logs from the right:

\(\displaystyle 4 = \ln x\)

\(\displaystyle e^4 = x\)

Which is one of the answers, the other being 1. Where did the one come from?

And the second, I'm not really even sure what to do.

\(\displaystyle (\log_{3} x)^2 - \log_{3} x^2 = 3\)

I keep ending up with:

\(\displaystyle 27x^2 = x^{\log_{3} x}\)

And from there I'm stuck.

Thanks for any help.

 

[Opalg]

I have two equations I can't figure out how to solve.

\(\displaystyle \sqrt{\ln x} = \ln \sqrt{x}\)

I started off squaring both sides:

\(\displaystyle \ln x = \frac{1}{4}(\ln x)(\ln x)\)

Then dividing by 1/4 and one of the logs from the right:

\(\displaystyle 4 = \ln x\)

\(\displaystyle e^4 = x\)

Which is one of the answers, the other being 1. Where did the one come from?

You lost it when you divided through by ln(x). You forgot to consider the possibility that ln(x) might be zero.

And the second, I'm not really even sure what to do.

\(\displaystyle (\log_{3} x)^2 - \log_{3} x^2 = 3\)

For the second problem, let \(\displaystyle y=\log_3x\). Then \(\displaystyle \log_3x^2=2y\), and you get the quadratic equation \(\displaystyle y^2-2y=3\).

 

[axrw]

Ok, so...

\(\displaystyle \ln x = \frac{1}{4}(\ln x)^2\)

Let \(\displaystyle u = \ln x\)

\(\displaystyle u^2 - 4u = 0\)

And then solving that I get the right answers of 1 and e^4.

And the second is just the same process. With the answers being 1/3 and 27.

Thanks, Opalg. I really appreciate your help.