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Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?
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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 ?
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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.
Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?
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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
Re: If (-5)^(4x) = 5^(9 + x) and x is an integer, what is the value of x ?
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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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96% (01:01) correct 
