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# If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4

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Math Expert
Joined: 02 Sep 2009
Posts: 64068
If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4  [#permalink]

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27 Mar 2020, 04:35
00:00

Difficulty:

65% (hard)

Question Stats:

63% (03:01) correct 37% (03:12) wrong based on 48 sessions

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If m and n are integers such that $$(\sqrt{2})^{19}*3^4*4^2*9^m*8^n = 3^n*16^m(\sqrt[4]{64})$$ then m is

A. -24
B. -20
C. -18
D. -16
E. -12

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Math Expert
Joined: 02 Aug 2009
Posts: 8584
If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4  [#permalink]

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27 Mar 2020, 20:11
2
Bunuel wrote:
If m and n are integers such that $$(\sqrt{2})^{19}*3^4*4^2*9^m*8^n = 3^n*16^m(\sqrt[4]{64})$$ then m is

A. -24
B. -20
C. -18
D. -16
E. -12

There are two variables - m and n.

$$(\sqrt{2})^{19}*3^4*4^2*9^m*8^n = 3^n*16^m(\sqrt[4]{64})$$....
$$2^{\frac{19}{2}}*3^4*(2^2)^2*3^{2m}*(2^3)^n = 3^n*2^{4m}(\sqrt[4]{2^6})..........2^{(\frac{19}{2}+4+3n)}*3^{4+2m}=2^{(\frac{6}{4}+4m)}*3^n$$

Equate the powers of 3..
$$4+2m=n$$
Now equate the powers of 2..
$$(\frac{19}{2}+4+3n)={\frac{6}{4}+4m}$$

substitute n=4+2m.....
$$(\frac{19}{2}+4+3*(4+2m))}={\frac{6}{4}+4m}...........6m-4m=\frac{3}{2}-(\frac{19}{2}+4+12).....2m=\frac{3-51}{2}=-24.....m=-12$$
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Re: If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4  [#permalink]

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28 Mar 2020, 01:05
Bunuel wrote:
If m and n are integers such that $$(\sqrt{2})^{19}*3^4*4^2*9^m*8^n = 3^n*16^m(\sqrt[4]{64})$$ then m is

A. -24
B. -20
C. -18
D. -16
E. -12

Please find the solution as explained in image below

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Screenshot 2020-03-28 at 2.34.29 PM.png [ 829.97 KiB | Viewed 480 times ]

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Re: If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4   [#permalink] 28 Mar 2020, 01:05