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 Post subject: algebra questionPosted: Mon Nov 23, 2009 12:26 pm

Joined: Sat Nov 14, 2009 6:39 pm
Posts: 1
hi!
could someone please explain question 2 on page 7 of the algebra book? thanks!

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 Post subject: Re: algebra questionPosted: Sat Dec 12, 2009 3:31 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Our first goal, any time we have variables in our exponents, is to get the bases the same on both sides of the equation.

Look first at 75^y. 75 is the same as 5*5*3, so let's make it 5^5 * 5^y * 3^y. Combining the 5's gives us 5^2y. Our entire term is now 3^y * 5^2y.
27 is the same as 3^3, so we can adjust this term to become 3^(6y+3)

Now, we have 3^y * 5^2y * 3^(6y+3) on the left.
Combine the 3's and we have 3^(7y+3) * 5^2y
At this point, our bases are the same on both sides, so we know that
2y=4, which means y=2
and
7y+3 = x
Since we know that y=2, we know 7y+3 is 17, and that x=17.

Hope this helps!

Veritas Help

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