Hello, AmandaLynn04!
I think you have a typo in your equation . . . I'll correct it.
Solve for \(\displaystyle x:\;\;e^{2x}\,+\,9e^{-2x}\:=\:6\)
Multiply by \(\displaystyle e^{2x}:\;\;e^{4x}\,+\,9\:=\:6e^{2x}\)
Then we have:
.\(\displaystyle e^{4x}\,-\,6e^{2x}\,+\,9\:=\:0\)
. . which factors:
.\(\displaystyle (e^{2x}\,-\,3)^2\:=\:0\)
Then:
.\(\displaystyle e^{2x}\,-\,3\:=\:0\;\;\Rightarrow\;\;e^{2x}\,=\,3\)
Take logs:
.\(\displaystyle \ln(e^{2x})\:=\:\ln(3)\;\;\Rightarrow\;\;2x\cdot\ln(e)\,=\,\ln(3)\;\;\Rightarrow\;\;2x\cdot1\,=\,\ln(3)\)
Therefore:
.\(\displaystyle 2x\,=\,\ln(3)\;\;\Rightarrow\;\;x\,=\,\frac{\ln(3)}{2}\)