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# For what value of x does (−6)^x = 6^(9−x)?

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Math Expert
Joined: 02 Sep 2009
Posts: 58316
For what value of x does (−6)^x = 6^(9−x)?  [#permalink]

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01 Mar 2017, 02:02
00:00

Difficulty:

45% (medium)

Question Stats:

65% (01:34) correct 35% (01:32) wrong based on 176 sessions

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For what value of x does $$(−6)^x = 6^{(9−x)}$$?

A. −9/2
B. 0
C. 1
D. 9/2
E. There is no real-number solution

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Senior Manager
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: For what value of x does (−6)^x = 6^(9−x)?  [#permalink]

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01 Mar 2017, 09:58
1
2
Bunuel wrote:
For what value of x does $$(−6)^x = 6^{(9−x)}$$?

A. −9/2
B. 0
C. 1
D. 9/2
E. There is no real-number solution

Hi

$$a^b = (-a)^b$$ only if b is even.

In our case, expression to the right is always positive. In order for the expresion to the left to be positive x should be even, but 9-x when x is even will always be odd.
On the other hand, 9/2 or -9/2 will bring us square root of a negative number - complex number.

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Re: For what value of x does (−6)^x = 6^(9−x)?  [#permalink]

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01 Mar 2017, 10:18
Answer is E, with options A,B,C the exponent part will never be equal on LHS and RHS....for option D the base on both sides are is equal only in mod, hence no real solution exists for the equation

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Senior Manager
Joined: 12 Sep 2017
Posts: 301
For what value of x does (−6)^x = 6^(9−x)?  [#permalink]

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14 Feb 2019, 20:51
Bunuel wrote:
For what value of x does $$(−6)^x = 6^{(9−x)}$$?

A. −9/2
B. 0
C. 1
D. 9/2
E. There is no real-number solution

Hello there math experts!!!

What's wrong with doing it like:

$$(−6)^x = 6^{(9−x)}$$ ... Let x be 9/2

$$(−6)^-(9/2) = 6^{(9−(9/2))}$$ is it possible to cancel out the minus from -6????

$$(6)^(9/2) = 6^{((18-9)/2))}$$

$$(6)^(9/2) = 6^{(9/2)}$$
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Re: For what value of x does (−6)^x = 6^(9−x)?  [#permalink]

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18 Feb 2019, 19:13
Bunuel wrote:
For what value of x does $$(−6)^x = 6^{(9−x)}$$?

A. −9/2
B. 0
C. 1
D. 9/2
E. There is no real-number solution

In order for the equation to be true, the exponents must be equal and they must be an even number such as 2 (notice that (-6)^2 = 6^2) or a fraction in lowest terms in the form of m/n where m is even and n is odd such as ⅔ (notice that (-6)^(⅔) = 6^(⅔)). However, setting the exponents equal, we find that:

x = 9 - x

2x = 9

x = 9/2

Since 9/2 is neither of the two cases mentioned above, there is no real-number solution that will allow -6^x to equal 6^(9-x).

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Re: For what value of x does (−6)^x = 6^(9−x)?   [#permalink] 18 Feb 2019, 19:13
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