# What is the logic behind this answer?

$$9^x = 2 \times 3^{x}+6$$

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 = $9^{xx}$)
• $\left(a^b\right)^c=a^{bc}=a^{cb}=\left(a^c\right)^b$ Dec 23, 2012 at 19:10
• Multiplication is commutative... Dec 23, 2012 at 19:11
• Your RHS is incorrect. You put down $9^{xx}$, which is $9^{(x^2)}$, not $(9^x)^2$. Dec 23, 2012 at 19:27
• I think Brian M. Scott's comment is more to the point than Fant's is. You can't use commutativity of multiplication until what you've got is multiplication. Dec 23, 2012 at 22:20

I am not sure what kind of answer you are looking for, but perhaps this will help: $$\left(3^{x}\right)^{2}=3^{x}\cdot3^{x}=3^{x+x}=3^{2x}$$ and $$\left(3^{2}\right)^{x}=\left(3\cdot3\right)^{x}=3^{x}\cdot3^{x}=3^{2x}.$$
In general, $$\left(x^y\right)^z=x^{yz}=(x^z)^y.$$