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

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 = $9^{xx}$)
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$\left(a^b\right)^c=a^{bc}=a^{cb}=\left(a^c\right)^b$ – Brian M. Scott Dec 23 '12 at 19:10
Multiplication is commutative... – user127.0.0.1 Dec 23 '12 at 19:11
Your RHS is incorrect. You put down $9^{xx}$, which is $9^{(x^2)}$, not $(9^x)^2$. – Joe Z. Dec 23 '12 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. – Michael Hardy Dec 23 '12 at 22:20
up vote 5 down vote accepted

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, $$\left(x^y\right)^z=x^{yz}=(x^z)^y.$$

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Note: This law only holds if the base/power are real... – apnorton Dec 23 '12 at 20:50

Consider the following theorems (with proofs): Power of Power and Power of Product. These results should help you understand the concepts better. Needless to say, real multiplication is commutative.

share|cite|improve this answer here is a pretty good collection of examples and explanations to your problem. it is a simple property of exponents.

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