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Solving For X

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Solving For X

by ajmoney09 » Wed Aug 03, 2011 11:13 am
Can Someone help with this please? The explanation skips a few steps that I don't understand so a nice little walk through will really help here!
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by Anurag@Gurome » Wed Aug 03, 2011 7:39 pm
ajmoney09 wrote:Can Someone help with this please? The explanation skips a few steps that I don't understand so a nice little walk through will really help here!
[15^x + (15^x * 15^1)]/(4^y) = 15^y
15^x[1 + 15] = 15^y * 4^y
15^x * 16 = 15^y * 4^y
15^x * 2^4 = 15^y * 2^(2y)
Since the bases are the same on left hand and right hand side, so exponents will also be the same.
So, x = y and 4 = 2y implies y = 2 and x = 2

The correct answer is A.
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by ajmoney09 » Thu Aug 04, 2011 10:17 am
Anurag@Gurome wrote:
ajmoney09 wrote:Can Someone help with this please? The explanation skips a few steps that I don't understand so a nice little walk through will really help here!
[15^x + (15^x * 15^1)]/(4^y) = 15^y
15^x[1 + 15] = 15^y * 4^y
15^x * 16 = 15^y * 4^y
15^x * 2^4 = 15^y * 2^(2y)
Since the bases are the same on left hand and right hand side, so exponents will also be the same.
So, x = y and 4 = 2y implies y = 2 and x = 2

The correct answer is A.
Thanks foryour help! Butim not sure how take the step below.

[15^x + (15^x * 15^1)]/(4^y) = 15^y
15^x[1 + 15] = 15^y * 4^y

I get that you factored out the 15^x but you only add 1 but there are 2 15^x values..... And not sure how you can factor out 15^x from 15^1
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