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Sub 505 (Easy)|   Exponents|      
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Bunuel
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 02 Sep 2015, 22:30
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B
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If \(8^{x}9^{2y}=81(2^{12y})\), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16


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B
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 02 Sep 2015, 22:55
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Bunuel
If \(8^{x}9^{2y}=81(2^{12y})\), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16


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2^3x*3^4y=3^4*2^12y
Equating powers with equal bases,
3x=12y &4y=4
y=1
Therefore, 3x=12
x=4
Answer B
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 03 Sep 2015, 01:03
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How is it possible to equate the bases ? They are not the same throughout the equation
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 03 Sep 2015, 08:58
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Bunuel
If \(8^{x}9^{2y}=81(2^{12y})\), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16


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\(8^{x}9^{2y}=81(2^{12y})\)

or \(8^{x}9^{2y}=9^2(8^{4y})\)

So,
2y=2
or y=1

Also,
x=4y
or x=4

Answer:-B
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 03 Sep 2015, 10:09
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noTh1ng
How is it possible to equate the bases ? They are not the same throughout the equation
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Good question. This is the reason you need to do the prime factorization of the "big" numbers

Recognise that 8 = \(2^3\)
9 = \(3^2\)
81=\(3^4\)

So now, you will have the bases either with 2s or 3s. Then equation the powers of 2s on either side of the equations and similarly equate the powers of 3s on either side.
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 03 Sep 2015, 10:55
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Solution:

change all bases into 2 and 3.
Then, we get 4y=4 and 3x= 12y.
So, y=1 and x=4.
So, answer is B
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 05 Sep 2015, 11:24
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Bunuel
If \(8^{x}9^{2y}=81(2^{12y})\), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16


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Making the same base on both the sides of equations we get (2^3x) x 9^2y = (2^12y) x 9^2

Equating the common bases,

1) 3x = 12y and 2) 2y = 2, there fore y =1 substituting in 1 we get x = 4

Correct answer is B
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 07 Sep 2015, 03:57
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Bunuel
If \(8^{x}9^{2y}=81(2^{12y})\), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16


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VERITAS PREP OFFICIAL SOLUTION:

In this problem, break down the larger numbers to find common bases for the exponents. \(8^{x}\) can be written as \((2^{3})^{x}\) and 81 can be written as \(9^{2}\), yielding \((2^{3})^{x} * 9^{2y} = 9^{2}(2^{12y})\). Because we now have common bases for the 9s, we can ignore \((2^3)^x\) and \(2^{12y}\) because 2 to any power does not share any factors with 9 to any power (since 9 = 3 x 3). Therefore, we can determine that 2y = 2, and that y = 1.

Plugging in for the 2 terms, we then have \((2^{3})^{x} = 2^{12}\).

As an exponential rule, \((2^{3})^{x}\) is equal to \(2^{3x}\), so we can again set the exponents with same bases equal to find that \(3x = 12\), and that \(x = 4\).

Answer: B.
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 07 Sep 2015, 04:29
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8^{x}9^{2y}=81(2^{12y})
(2^3x)(9^2y)=(9^2)(2^12y)
(2^3x)(9^2y)=(2^12y)(9^2)

As we know if the bases are same then EXPONENTS are equal. Therefore,
Base '9' EXPONENTS i.e 2y=2,
we get y=1


Similarly, For base '2'

3x=12y
x=4y
but we got y=1 therefore, x=4

Hence Answer is B
:-D
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If 8^{x}9^{2y}=81(2^{12y}), what is the value of x? 27 Dec 2020, 02:43
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