## What are the steps to solving this problem?

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rogermiranda
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### What are the steps to solving this problem?

I have 3 raised to the (x^2 - 1) / 9 raised to (x + 2) = one-nineth.

So first I set up common bases.
I have 3 raised to the (x^2 - 1) divided by 3 raised to the (2x + 4) = one-nineth.
Now I am stuck. I have the same bases but am I suppose subtract the exponents?

stapel_eliz
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rogermiranda wrote:I have 3 raised to the (x^2 - 1) / 9 raised to (x + 2) = one-nineth

Does your formatting mean the following?

. . . . .$\frac{3^{x^2 - 1}}{9^{x + 2}}\, =\, \frac{1}{9}$

If so, then...

rogermiranda wrote:So first I set up common bases.

What must be the value of $a$, for $3^a\, =\, 1?$ What must be the value of $b$, for $3^b\, =\, 9?$

By using this information to convert the right-hand side to a fraction in powers of 3 (or alternatively, using exponent rules to convert each side to possibly negative powers of 3), you can set up an equivalence that you can solve for the value(s) of $x.$

rogermiranda
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### Re: What are the steps to solving this problem?

I'm sorry, I still cannot sovle this equation.
I followed your advice and converted the right side to equal bases.
This is what I now have,
3 raised to (x^2 - 1) = 3^0
3 raised to (2x + 4) = 3^2

So I have x^2 - 1 = 0 and 2x + 4 = 2. So I solved both equations. x= 1 and x = -1
My textbook has the answer x= 3 or x = - 1. But where does the 3 come from?

stapel_eliz
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I hadn't worked out all the steps to the end earlier (sorry!), but checking your solution "x = +1" would have shown that this value does not work.

Using the other method outlined (longer and a bit more complicated, but generally more reliable), you will end up with:

. . . . .$3^{\left(x^2 -2x - 5\right)}\, =\, 3^{-2}$

This gives you a quadratic to solve, whose solutions (upon checking) do work, and also match the book.