What are the steps to solving this problem?  TOPIC_SOLVED

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What are the steps to solving this problem?

Postby rogermiranda on Tue Jun 15, 2010 12:40 pm

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?
Can someone please help me get over this hump?
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Postby stapel_eliz on Tue Jun 15, 2010 5:28 pm

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

Does your formatting mean the following?

. . . . .

If so, then...

rogermiranda wrote:So first I set up common bases.

What must be the value of , for What must be the value of , for

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 :wink:
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Re: What are the steps to solving this problem?

Postby rogermiranda on Wed Jun 16, 2010 12:11 am

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?
What am I doing wrong? Please help.
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  TOPIC_SOLVED

Postby stapel_eliz on Wed Jun 16, 2010 11:24 am

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:

. . . . .

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