What is the logic behind this answer?

Question

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

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 = $9^{xx}$)

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

score 5 · Accepted Answer · edited Dec 23 '12 at 19:39

5

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

edited Dec 23 '12 at 19:39

Joe Z.

6,719

answered Dec 23 '12 at 19:14

Eric Naslund

72,099

2

Note: This law only holds if the base/power are real... http://mathworld.wolfram.com/ExponentLaws.html – apnorton Dec 23 '12 at 20:50

score 1 · Answer 2 · answered Dec 23 '12 at 19:31

1

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.

answered Dec 23 '12 at 19:31

glebovg

10,154

score 1 · Answer 3 · answered Dec 23 '12 at 19:52

1

http://hotmath.com/hotmath_help/topics/properties-of-exponents.html here is a pretty good collection of examples and explanations to your problem. it is a simple property of exponents.

answered Dec 23 '12 at 19:52

doniyor

3,700

What is the logic behind this answer?

3 Answers3