Solve please

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Solve please

by Hmna » Mon May 08, 2017 9:25 am
For the equation a^x/a^y=a^-10 and (a^y)^3=a^-x, if a>1 what is the value of x

A)2.5
B)-7.5
C)2
D)1

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by Brent@GMATPrepNow » Mon May 08, 2017 12:05 pm
Hmna wrote:For the equation a^x/a^y = a^-10 and (a^y)^3 = a^-x, if a>1 what is the value of x

A)2.5
B)-7.5
C)2
D)1
Given: a^x/a^y = a^(-10)
Simplify: a^(x-y) = a^(-10)
So, we can conclude that x-y = -10

Given: (a^y)^3 = a^-x
Simplify: a^(3y) = a^(-x)
So, we can conclude that 3y = -x

We have two equations:
x-y = -10
3y = -x

Take x-y = -10 and solve for y to get: y = x + 10
Now take 3y = -x and replace y with x + 10 to get: 3(x + 10) = -x
Expand: 3x + 30 = -x
Subtract 3x from both sides: 30 = -4x
Solve: x = -7.5

Answer: B
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by GMATGuruNY » Mon May 08, 2017 1:01 pm
Hmna wrote:For the equation a^x/a^y=a^-10 and (a^y)^3=a^-x, if a>1 what is the value of x

A)2.5
B)-7.5
C)2
D)1
The second equation can be rephrased as a^(3y) = a^(-x).

Dividing the rephrased second equation by the first, we get:
a^(3y) / (a^x/a^y) = a^(-x) / a¯¹�

a^(3y) * (a^y/a^x) = a¹�/a^x

a^(4y) = a¹�

4y = 10

y = 10/4 = 2.5.

Substituting y = 2.5 into a^(3y) = a^(-x), we get:
a^(3 * 2.5) = a^(-x)

a^(7.5) = a^(-x)

7.5 = -x

x = -7.5.

The correct answer is B.
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