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by **maroonblazer** on Sat Nov 26, 2011 6:05 pm

The book asks:
If $25^{b+1}= x$ and
$5^b = y$

then express x in terms of y.

The answer it gives is:

$x=25y^2$

I'm struggling to understand the formula or operation(s) one would go through to answer this.

Help?
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by **maggiemagnet** on Sat Nov 26, 2011 11:08 pm

maroonblazer wrote:If $25^{b+1}= x$ and
$5^b = y$

then express x in terms of y.

Try rewriting 25^b+1 so the base is 5. If you can it so it's all in terms of 5^b, then you can plug "y" in, and see what you get.
:clap:

by **maroonblazer** on Sun Nov 27, 2011 12:25 am

Ok, so that would be:
$25^{b+1} = 5^{b+2}$ , right?

Still stumped...

by **maggiemagnet** on Sun Nov 27, 2011 10:45 pm

maroonblazer wrote:Ok, so that would be:
$25^{b+1} = 5^{b+2}$ , right?

Still stumped...

How did you get that? Try showing all the steps:

25^b+1 = (5²)^b+1 = 5^2(b+1) = ...

:clap:

by **pulsar_qsr** on Sat Dec 10, 2011 8:02 pm

25^(b+1) = x
and 5^b = y
or, 5^(b+1) = 5y
or, 5^2(b+1) = 5y^2
or, 25^(b+1) = 25y^2
or, x = 25y^2

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