$9^x = 2 \times 3^{x}+6$
The method in my book to solve this:
$(3^2)^x = 2 \times 3^x + 6$
$(3^x)^2 = 2 \times 3^x + 6$
$ p=3^x, p^2=2p+6$
After using quadratic equation we get the answers ($x = \log_3(1+\sqrt{7})$)
This bothers me:
- Why does $(3^2)^x = (3^x)^2$, this seems incorrect to me (LHS = $9^x$ and RHS = $9^{xx}$)