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655-705 (Hard)|   Algebra|   Exponents|   Roots|      
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Bunuel
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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 27 Mar 2020, 05:35
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E
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Difficulty:

65% (hard)

Question Stats:

69% (03:15) correct 31% (03:06) wrong based on 493 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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E
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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 27 Mar 2020, 21:11
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Bunuel
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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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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If m and n are integers such that (√2)^19*3^4*4^2*9^m*8^n = 3^n*16^m(4 28 Mar 2020, 02:05
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Bunuel
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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Please find the solution as explained in image below

Answer: Option E
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Screenshot 2020-03-28 at 2.34.29 PM.png
Screenshot 2020-03-28 at 2.34.29 PM.png [ 829.97 KiB | Viewed 9909 times ]

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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 21 Mar 2024, 23:44
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4+2m=n ...(1)

9.5+4+3n=4m+1.5 ...(2)
12+3n=4m

=>12+12+6m=4m
2m=-24
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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 25 Jun 2024, 08:15
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\((\sqrt{2})^{19} = 2^9\sqrt{2}\)
\(\sqrt[4]{64} = 2^{6/4} = 2^1*2^{1/2}\)

==> Now group and reduce bases: 

\(2^{3n} * 3^{2m+4} * 2^4 * 2^9 *2^{1/2} = 2^1 * 2^{1/2} * 2^{4m}* 3^n\)­

==> By comparing base 3 on both sides we get: \(2m + 4 = n\) ---- eq 1

==> Remove known common exponents \((2^1 and  2^{1/2})\) from both the sides:

\(2^{3n} * 3^{2m+4} * 2^4 * 2^8 = 2^{4m}* 3^n\)­

==> We then get our second equation: \(4m = 3n + 12\) ----- eq 2

Solve eq 1 & 2 using substitution to get \(m = -12\)­
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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 02 Jan 2026, 13:32
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