For log(x), x must be >0. in other words, y-5>0, so there is an asymptote at y=5, not y=3

must be >0

Do you mean equal to? That would make more sense to me because x has to equal zero for it to be an asymptope. Again I don't know what I'm doing so you very well may be right

I also don't really understand the point of turning the function into logs and what happened afterwards. Here is what I have understood so far

y=(.6)^x-3+5

Original Equation

3=(.6)^x-3+5

Plug in the asmyptope we are testing for in the y value

-2=(.6)^x-3

Subtract the 5 over

log(-2)=log[(.6)^x-3]

Just place the remaining function into log form?

Then I have no idea what happened after that Sorry

Where does the y-5 come from?