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GMATGuruNY
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Plug in a value for x and solve for y.
Let x=3, so that 5^(x-3) = 5� =1.
Plugging x=3 into the equation, we get:
5³ - 5� = 124 * 5^y
124 = 124 * 5^y
1 = 5^y
y=0.
Since the question stem asks for the value of y, our target is y=0.
Now plug x=3 into the answers to see which yields our target of 0.
A quick scan of the answer choices shows that only C works:
x-3 = 3-3 = 0.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Brent@GMATPrepNow
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One option is to rewrite the left side of the equation by factoring out 5^(x-3)If 5^x - 5^(x-3) = (124)(5^y), what is y in terms of x?
A) x
B) x - 6
C) x - 3
D) 2x + 3
E) 2x + 6
So, we get: 5^(x-3)[5^3 - 1] = (124)(5^y)
Evaluate to get: 5^(x-3)[125 - 1] = (124)(5^y)
Simplify to get: 5^(x-3)[124] = (124)(5^y)
Divide both sides by 124 to get: 5^(x-3) = 5^y
So, x-3 = y
Answer: C
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ASIDE: A lot of students struggle to see how we can factor 5^x - 5^(x-3) to get 5^(x-3)[5^3 - 1]
Sure, they may be okay with straightforward factoring like these examples:
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)
But they have problems when the exponents are variables.
IMPORTANT: Notice that, each time, the greatest common factor of both terms is the term with the smaller exponent.
So, in the expression 5^x - 5^(x-3), the term with the smaller exponent is 5^(x-3, so we can factor out 5^(x-3)
Likewise, w^x + x^(x+5) = w^x(1 + w^5)
And 2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]
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