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If \(9^{2x+5}=27^{3x-10},\) then \(x =\)

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Re: If \(9^{2x+5}=27^{3x-10},\) then \(x =\)

by Chandanagarwal » Thu Oct 01, 2020 9:08 am
2(2x+5) = 3(3x-10)

4x+10 = 9x -30

5x = 40

x=8
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Re: If \(9^{2x+5}=27^{3x-10},\) then \(x =\)

by psarma » Thu Oct 01, 2020 3:27 pm
\(9^{2x+5}\) = \(3^{2\left(2x+5\right)}\)
= \(3^{4x+10}\)
Likewise,
\(27^{3x-10}\) = \(3^{3\left(3x-10\right)}\)
= \(3^{9x-30}\)

Equating the powers together;
4x+10 = 9x-30
i.e 5x=40
or x=8
Answer C
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Re: If \(9^{2x+5}=27^{3x-10},\) then \(x =\)

by Scott@TargetTestPrep » Tue Oct 06, 2020 3:18 am
Gmat_mission wrote: ↑
Thu Oct 01, 2020 7:04 am
 If \(9^{2x+5}=27^{3x-10},\) then \(x =\)

A. 3
B. 6
C. 8
D. 12
E. 15

Answer: C

Solution:

We first must get a base of 3 for each expression so that we can then equate the exponents. Simplifying, we have:

(3^2)^(2x + 5) = (3^3)^(3x - 10)

3^(4x + 10) = 3^(9x - 30)

4x + 10 = 9x - 30

40 = 5x

8 = x

Answer: C

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