Attached is the equation. Could someone show me the steps? I know you had to cross-multiply something.
Is it x = 2x - 1?
Please get the concept of "cross multiply" out of your life. It is not your friend. The fact that it doesn't even remotely apply to this problem suggests to me that it is as confusing as I like to say it is.
You almost have this one, but failed to spot that the bases were different. If they were both 8 or both 16, you would be right on. You must change to a common base.
$\displaystyle 16 = 2^{4}$ and $\displaystyle 8 = 2^{3}$
Use those hints and show us what you get.
Its actually a problem that is very similar to the first one. Whenever you see a power on one side and some number on the other, always take logs on both sides. And then try to play with the logs.
For example:
$\displaystyle 16^x = 8^{2x-1} \Rightarrow x \log 16 = (2x - 1) \log 8 $$\displaystyle \Rightarrow x (4\log 2) = (2x - 1) (3\log 2) \Rightarrow 4x = 3(2x-1)$
Similarly:
$\displaystyle e^{\ln (x+2)} = 6 \Rightarrow \ln e^{\ln (x+2)} = \ln 6 \Rightarrow$$\displaystyle \ln (x+2) \ln e = \ln 6 \Rightarrow \ln (x+2) = \ln 6 \Rightarrow x+2 = 6$
This method is very mechanical. Identifying both sides as powers of the same base is faster, if you get used to it. But you will need practice before you can do that.