Problem Solving

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

One option is to rewrite the left side of the equation by factoring out 5^(x-3)
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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Cheers,
Brent

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

Another option is to PLUG IN a value for x and see what kind of relationship we get between x and y.
There are two "nice" x-values to plug in. They are x = 0 and x = 3, since we can easily use these to evaluate 5^x and 5^(x-3). Of these two values, x = 3 is the easier one to plug in.
So, let's plug in x = 3
We get: 5^3 - 5^(3-3) = (124)(5^y)
Simplify to get: 125 - 1 = (124)(5^y)
Simplify to get: 124 = (124)(5^y)
Divide both sides by 124 to get: 1 = 5^y
Solve for y to get: y = 0

So, when x = 3, y = 0.

Now we'll check the answer choices to see which one satisfies this relationship.
A) y = x... So, we get 0 = 3 (NOPE)
B) y = x - 6... So, we get 0 = 3 - 6 (NOPE)
C) y = x - 3... So, we get 0 = 3 - 3 IT WORKS!
D) y = 2x + 3... So, we get 0 = 2(3) + 3 (NOPE)
E) y = 2x + 6... So, we get 0 = 2(3) + 6 (NOPE)

Answer: C

Cheers,
Brent

Here are two more practice questions where we have variable exponents:

https://www.beatthegmat.com/is-5-k-less- ... 74188.html
https://www.beatthegmat.com/question-on- ... 68718.html


Cheers,
Brent