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If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?

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If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 06 Dec 2015, 10:55
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 06 Dec 2015, 19:27
1
Since x is an integer, (-5)^4x is always positive.
So, 5^4x = 5^(9 + x)
4x = 9 + x
3x = 9
x = 3

Answer: C
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 06 Dec 2015, 23:25
since the values are equal
the powers can be equated.
we therefore have

4x=9-x
x=3
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 07 Dec 2015, 00:02
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?

A. 5
B. 4
C. 3
D. 2
E. 1


Since the bases of both sides are -5 and 5, the exponents of both sides should be even numbers(i. e. 4x and 9+x should be even numbers).

Moreover they should be equal since the absolute values of the bases are equal.
That is 4x=9+x ---> x=3. The answer is, therefore, C.
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 09 Dec 2015, 07:51
Since (-5)^4x = 5^(9+x)

Thus both sides are equal: Hence x>0

4x = 9+x
x = 3

Ans: 3
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 29 Oct 2016, 08:28
(-5)^4x = 5^(9+x)

Negative on LHS is not needed as any multiple of 4 will be an even integer, canceling out the negative, thus we can drop it and treat it as 5. Since both sides will now have 5 as a common base, we can equate the exponents and solve.

4x=9+x
3x=9
x=3
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 29 Oct 2016, 08:48
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Bunuel wrote:
If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?

A. 5
B. 4
C. 3
D. 2
E. 1

The important concept here is: (negative number)^(even integer) = some positive value
Examples
(-5)^2 = 25
(-3)^4 = 81
(-1)^10 = 1
(-2)^6 = 64

Also notice that:
(-5)^2 = (5)^2 = 25
(-3)^4 = (3)^4 = 81
(-1)^10 = (1)^10 = 1
(-2)^6 = (2)^6 = 64
The same can be said for ANY negative value raised to an EVEN power

Now onto the question.............
Since 4x is EVEN for any integer x, we can write:
(-5)^(4x) = (5)^(4x)

So, we can write: 5^(4x) = 5^(9 + x)
Since we now have the base, we can conclude that 4x = 9 + x
Solve to get: x = 3

Answer:

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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 29 Oct 2016, 09:59
Bunuel wrote:
If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?

A. 5
B. 4
C. 3
D. 2
E. 1


(-5)^(4x) = 5^(9 + x) = -1(4x)*(5)^(4x) = 5^(9 + x)

Or, 4x = 9 + x

Or, 3x = 9

Or, x = 3

Hence answer will be (C) 3

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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 17 Dec 2016, 10:52
simple and good question
First few seconds I tried solving as\(\frac{1}{5^{4x}}\)
Later, I substituted options
Any negative value raised to an Even power is always become Positive value
Thanks for explanations
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?  [#permalink]

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New post 09 Feb 2020, 04:21
Bunuel wrote:
If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?

A. 5
B. 4
C. 3
D. 2
E. 1


Since x is an integer, 4x is even, and (-5)^(4x) = 5^(4x). Thus, we can drop the bases, equate the exponents, and solve for x.

4x = 9 + x

3x = 9

x = 3

Answer: C
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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?   [#permalink] 09 Feb 2020, 04:21
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