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If 9^y + 3^b = 10*3^b, then 2y ='

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by DavidG@VeritasPrep » Thu Mar 01, 2018 12:48 pm
Roland2rule wrote:If 9^y + 3^b = 10*3^b, then 2y ='

(A) b - 2
(B) b - 1
(C) b
(D) b + 1
(E) b + 2
OA is E
Please Expert, can you help me out with this problem<i class="em em-expressionless"></i>? Thanks
This could have been depicted a little more cleanly. Let's assume the initial equation is 9^y + 3^b = 10 * (3^b)
Say b = 1. We get 9^y + 3^1 = 10 * (3^1)
9^y = 27

3^(2y) = 3^3
2y = 3. 3 is our target.

When we plug 1 in for b, only E gives us 3.
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by regor60 » Fri Mar 02, 2018 11:08 am
Roland2rule wrote:If 9^y + 3^b = 10*3^b, then 2y ='

(A) b - 2
(B) b - 1
(C) b
(D) b + 1
(E) b + 2
OA is E
Please Expert, can you help me out with this problem<i class="em em-expressionless"></i>? Thanks
Or you could factor the thing:

9^y = (3^2)^y = 3^2y

Rewriting: 3^2y + 3^b =10(3^b)

Factor the left side: (3^b)(3^(2y-b) + 1), so

(3^b)(3^(2y-b) + 1) = 10(3^b)

therefore 3^(2y-b)+1 = 10

and 3^(2y-b) = 9

since 3^2 = 9 , 2y - b must equal 2, therefore

2y-b=2 and 2y = b+2, E
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by GMATGuruNY » Fri Mar 02, 2018 2:49 pm
Roland2rule wrote:If 9^y + 3^b = 10*3^b, then 2y ='

(A) b - 2
(B) b - 1
(C) b
(D) b + 1
(E) b + 2
So that all of the terms with exponents have the same base, let b=2.

Plugging b=9 into 9^y + 3^b = 10(3^b), we get:
9^y + 3² = 10(3²)
9^y + 9 = 10*9
9^y + 9 = 90
9^y = 81
y =2, with the result that 2y = 2*2 = 4.

The question stem asks for the value of 2y (4).
Now plug b=2 into the answer choices to see which yields the target value of 4.
Only E works:
b+2 = 2+2 = 4.

The correct answer is E.
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by [email protected] » Fri Mar 02, 2018 6:01 pm
Hi Roland2rule,

TESTing VALUES works perfectly on this question (as Mitch showed in his explanation). Since there are no 'restrictions' on the values of B and Y (and we're dealing with exponents), you can use the simplest value possible for B... B=0

IF... B=0, then we have...
9^Y + 3^0 = (10)(3^0)
9^Y + 1 = 10(1)
9^Y = 9
Y = 1

We're asked for the value of 2Y, so we're looking for an answer that equals 2(1) = 2 when B=0. There's only one answer that matches.

Final Answer: E

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by Scott@TargetTestPrep » Tue Mar 06, 2018 9:19 am
Roland2rule wrote:If 9^y + 3^b = 10*3^b, then 2y ='

(A) b - 2
(B) b - 1
(C) b
(D) b + 1
(E) b + 2
We see that 9^y = (3^2)^y. Simplifying the equation, we have:

3^2y + 3^b = 10(3^b)

3^2y = 10(3^b) - 3^b

Factoring out 3^b from both terms on the right side of the equation yields:

3^2y = 3^b(10 - 1)

3^2y = 3^b(9)

3^2y = 3^b(3^2)

3^2y = 3^(b + 2)

2y = b + 2

Answer: E

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