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# 9^k*27^(2k) =

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Math Expert
Joined: 02 Sep 2009
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23 Oct 2018, 21:53
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Difficulty:

5% (low)

Question Stats:

92% (00:38) correct 8% (00:59) wrong based on 59 sessions

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$$9^k*27^{2k} =$$

A. $$3^{5+3k}$$

B. $$3^{8k}$$

C. $$3^{11k}$$

D. $$3^{12k}$$

E. $$3^{12k^2}$$

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23 Oct 2018, 22:29
3^2k * 3^6k

3^(2k+6k)

3^8k

OPTION : B
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Joined: 23 Oct 2018
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GMAT 1: 760 Q50 V42

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23 Oct 2018, 23:06
1
Pretty straightforward question if we remember our exponent properties.

$$((a^b)^c) = a^{bc}$$
$$(a^b * a^c) = a^{b+c}$$

Applying here...
$$9^k∗27^{2k}=$$
$$(3^2)^k∗(3^3)^{2k}=$$
$$3^{2k}*(3^{6k})=$$
$$3^{8k}$$

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23 Oct 2018, 23:10
Bunuel wrote:
$$9^k*27^{2k} =$$

A. $$3^{5+3k}$$

B. $$3^{8k}$$

C. $$3^{11k}$$

D. $$3^{12k}$$

E. $$3^{12k^2}$$

$$9^k*27^{2k}$$

Or, $$3^{2k}*3^{6k}$$

Or, $$3^{8k}$$, Answer must be (B)
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24 Oct 2018, 03:57

Solution

Given:
• $$9^k * 27^{2k}$$

To find:
• Which of the given answer choices is equal to the given expression

Approach and Working:
• $$9^k * 27^{2k} = 3^{2k} * 3^{3*2k} = 3^{2k + 6k} = 3^{8k}$$

Hence the correct answer is Option B.

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24 Oct 2018, 05:06
Bunuel wrote:
$$9^k*27^{2k} =$$

A. $$3^{5+3k}$$

B. $$3^{8k}$$

C. $$3^{11k}$$

D. $$3^{12k}$$

E. $$3^{12k^2}$$

$$9^k 27^{2k}$$

$$3^{2k} 3^{6k}$$

$$3^{8k}$$

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25 Oct 2018, 17:59
Bunuel wrote:
$$9^k*27^{2k} =$$

A. $$3^{5+3k}$$

B. $$3^{8k}$$

C. $$3^{11k}$$

D. $$3^{12k}$$

E. $$3^{12k^2}$$

The key to this problem’s solution is to re-express each of the numbers such that the base is 3.

We see that 9^k = (3^2)^k = 3^2k and 27^2k = (3^3)^2k = 3^6k

Thus, 9^k x 27^2k = 3^2k x 3^6k = 3^8k

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Re: 9^k*27^(2k) =   [#permalink] 25 Oct 2018, 17:59
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