[GMAT math practice question]
Which of the following is equal to $$^{2^{k-1}3^{k+1}}?$$
$$A.\ 2^2\left(6^{k-1}\right)$$
$$B.\ 3^2\left(6^{k-1}\right)$$
$$C.\ 6\left(2^{k-1}\right)$$
$$D.\ 6\left(3^{k-1}\right)$$
$$E.\ 6^{k-1}$$
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Which of the following is equal to 2^{k-1}3^{k+1}?
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Max@Math Revolution
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Let k=1:Max@Math Revolution wrote:[GMAT math practice question]
Which of the following is equal to $${2^{k-1}3^{k+1}}?$$
$$A.\ 2^2\left(6^{k-1}\right)$$
$$B.\ 3^2\left(6^{k-1}\right)$$
$$C.\ 6\left(2^{k-1}\right)$$
$$D.\ 6\left(3^{k-1}\right)$$
$$E.\ 6^{k-1}$$
$${2^{k-1}3^{k+1}} = {2^{1-1}3^{1+1}} = {2^{0}3^{2}} = 1*9 = 9.$$
Now plug k=1 into the answers to see which yields a value of 9.
Only B works:
3²6^(k-1) = 9*6� = 9*1 = 9.
The correct answer is B.
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Option A:
$$2^2\left(6^{k-1}\right)=2^2\cdot\left(2^{k-1}\cdot3^{k-1}\right)=2^{k+1}\cdot3^{k+1}$$
Option B:
$$3^2\left(6^{k-1}\right)=3^2\cdot\left(2^{k-1}\cdot3^{k-1}\right)=2^{k-1}\cdot3^{k+1}$$
Option C:
$$6\left(2^{k-1}\right)=3\cdot2\cdot\left(2^{k-1}\right)=3\cdot2^k$$
Option D:
$$6\left(3^{k-1}\right)=3\cdot2\cdot\left(3^{k-1}\right)=2\cdot3^k$$
Option E:
$$6^{k-1}=2^{k-1}\cdot3^{k-1}$$
Option B is the correct answer. Regards!
$$2^2\left(6^{k-1}\right)=2^2\cdot\left(2^{k-1}\cdot3^{k-1}\right)=2^{k+1}\cdot3^{k+1}$$
Option B:
$$3^2\left(6^{k-1}\right)=3^2\cdot\left(2^{k-1}\cdot3^{k-1}\right)=2^{k-1}\cdot3^{k+1}$$
Option C:
$$6\left(2^{k-1}\right)=3\cdot2\cdot\left(2^{k-1}\right)=3\cdot2^k$$
Option D:
$$6\left(3^{k-1}\right)=3\cdot2\cdot\left(3^{k-1}\right)=2\cdot3^k$$
Option E:
$$6^{k-1}=2^{k-1}\cdot3^{k-1}$$
Option B is the correct answer. Regards!
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Scott@TargetTestPrep
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Recall that a^n x b^n = (ab)^n. That is, if the exponents are the same, we can multiply the bases.Max@Math Revolution wrote:[GMAT math practice question]
Which of the following is equal to $$^{2^{k-1}3^{k+1}}?$$
$$A.\ 2^2\left(6^{k-1}\right)$$
$$B.\ 3^2\left(6^{k-1}\right)$$
$$C.\ 6\left(2^{k-1}\right)$$
$$D.\ 6\left(3^{k-1}\right)$$
$$E.\ 6^{k-1}$$
Although here we don't have the same exponents (k - 1 vs. k + 1), we can "extract" the small exponent from the larger one, as follows:
2^(k-1) x 3^(k+1) = 2^(k-1) x 3^(k-1) x 3^2 = (2 x 3)^(k-1) x 3^2 = 6^(k-1) x 3^2
Answer: B
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Max@Math Revolution
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=>
$$2^{^{k-1}}3^{k+1}=2^{k-1}3^{k-1}\cdot3^2=3^2\cdot\left(2\cdot3\right)^{k-1}=3^2\left(6^{k-1}\right)$$
Therefore, the answer is B.
Answer: B
$$2^{^{k-1}}3^{k+1}=2^{k-1}3^{k-1}\cdot3^2=3^2\cdot\left(2\cdot3\right)^{k-1}=3^2\left(6^{k-1}\right)$$
Therefore, the answer is B.
Answer: B
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