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Exponents and roots

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Source: — Problem Solving |

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by [email protected] » Tue Nov 19, 2013 1:05 am
Hi Bharat,

We start off with the given equation (and we're told that x and y are integers, which will matter at the end of the question):

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

We have to do some math to clean everything up:

15^x + 15^(x+1) = (15^y)(4^y)

Let's factor out 15^x

15^x(1 + 15^1) = (15^y)(4^y)
15^x(16) = (15^y)(4^y)

The "16" on the left has to accounted for by the (4^y) on the right.

So y = 2

Then 15^x = 15^y

So x = 2

Final Answer: A

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by GMATGuruNY » Tue Nov 19, 2013 4:52 am
This question was posted previously with the answer choices shown here:
If x and y are integers and 15^x + 15^(x+1)/ 4^y = 15^y, what is the value of x?

a. 2
b. 3
c. 4
d. 5
e. Cannot be determined
15^x + 15^(x+1) = (15^y)(4^y).

We can plug in the answers, which represent the value of x.

Answer choice C: 4
15� + 15� = (15^y)(4^y)

15�(1+15) = (15^y)(4^y)

(15�)(4²) = (15^y)(4^y).

The values in red imply that y=2.
Thus, on the left-hand side, 15� needs to decrease to 15², implying that the value of x must be 2 LESS than answer choice C.

The correct answer is A.

Answer choice A: 2
15² + 15³ = (15^y)(4^y)

15²(1+15) = (15^y)(4^y)

(15²)(4²) = (15^y)(4^y).
Success!
Last edited by GMATGuruNY on Tue Nov 19, 2013 1:52 pm, edited 1 time in total.
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by Mathsbuddy » Tue Nov 19, 2013 7:19 am
15^x + 15^(x+1)/ 4y = 15y

15^x * (1 + 15) = 60y^2

15^x = 60/16 * y^2 = 15/4 * y^2 = (SQRT(15/4) * y)^2

If 15^x = p^2 where p = SQRT(15/4) * y

then when p = 15, x = 2.

However, this is only if y is an integer (as per the rules of the question):

p = 15 = SQRT(15/4) * y

so y^2 = 4 * 225/15 = 60

so y = SQRT(60) which is non-integer.

Therefore answer E - cannot be determined.
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by Mathsbuddy » Tue Nov 19, 2013 7:22 am
GMATGuruNY wrote:This question was posted previously with the answer choices shown here:
If x and y are integers and 15^x + 15^(x+1)/ 4y = 15y, what is the value of x?

a. 2
b. 3
c. 4
d. 5
e. Cannot be determined
15^x + 15^(x+1) = (15^y)(4^y).

We can plug in the answers, which represent the value of x.

Answer choice C: 4
15� + 15� = (15^y)(4^y)

15�(1+15) = (15^y)(4^y)

(15�)(4²) = (15^y)(4^y).

The values in red imply that y=2.
Thus, on the left-hand side, 15� needs to decrease to 15², implying that the value of x must be 2 LESS than answer choice C.

The correct answer is A.

Answer choice A: 2
15² + 15³ = (15^y)(4^y)

15²(1+15) = (15^y)(4^y)

(15²)(4²) = (15^y)(4^y).
Success!
Note however that if you expand this you get:
3600 = 60 * y^2
so y^2 = 60
but y = SQRT(60)is non-integer (contrary to the condition in the question).

Therefore answer E - cannot be determined!

I've made an algebraic solution to match this on a separate reply, if you are interested.
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by Mathsbuddy » Tue Nov 19, 2013 7:28 am
Mathsbuddy wrote:
GMATGuruNY wrote:This question was posted previously with the answer choices shown here:
If x and y are integers and 15^x + 15^(x+1)/ 4y = 15y, what is the value of x?

a. 2
b. 3
c. 4
d. 5
e. Cannot be determined
15^x + 15^(x+1) = (15^y)(4^y).

We can plug in the answers, which represent the value of x.

Answer choice C: 4
15� + 15� = (15^y)(4^y)

15�(1+15) = (15^y)(4^y)

(15�)(4²) = (15^y)(4^y).

The values in red imply that y=2.
Thus, on the left-hand side, 15� needs to decrease to 15², implying that the value of x must be 2 LESS than answer choice C.

The correct answer is A.

Answer choice A: 2
15² + 15³ = (15^y)(4^y)

15²(1+15) = (15^y)(4^y)

(15²)(4²) = (15^y)(4^y).
Success!
Note however that if you expand this you get:
3600 = 60 * y^2
so y^2 = 60
but y = SQRT(60)is non-integer (contrary to the condition in the question).

Therefore answer E - cannot be determined!

I've made an algebraic solution to match this on a separate reply, if you are interested.
I just found my own mistake. I can't believe it!
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by Mathsbuddy » Tue Nov 19, 2013 7:30 am
Mathsbuddy wrote:15^x + 15^(x+1)/ 4y = 15y

15^x * (1 + 15) = 60y^2

15^x = 60/16 * y^2 = 15/4 * y^2 = (SQRT(15/4) * y)^2

If 15^x = p^2 where p = SQRT(15/4) * y

then when p = 15, x = 2.

However, this is only if y is an integer (as per the rules of the question):

p = 15 = SQRT(15/4) * y

so y^2 = 4 * 225/15 = 60

so y = SQRT(60) which is non-integer.

Therefore answer E - cannot be determined.
I know I've made a mistake here, because y should equal 2.

Can anyone help me find my stupidity??? :)
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by [email protected] » Tue Nov 19, 2013 1:47 pm
Hi mathsbuddy,

Your "mistake" is probably dependent on which question you transcribed. The original question in the thread refers to (4^y) and (15^y). Mitch's reference question has typos is in (it uses 4y and 15y, but it's SUPPOSED to be 4^y and 15^y).

Your math was based on the wrong inputs.

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by GMATGuruNY » Tue Nov 19, 2013 2:06 pm
Yes, the question prompt that I copied into my post above contained typos. Rich, thanks for pointing them out. I've made the necessary corrections.
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