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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 = $9^{xx}$)
  • 1
    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.2012-12-23

3 Answers 3

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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... http://mathworld.wolfram.com/ExponentLaws.html2012-12-23
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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.

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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.