I added some brackets to avoid ambiguity.

melanie.espeland wrote: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

The left side can be factored to get: 5^(x-3)[5^3 - 1] = (124)(5^y)
Simplify the left side: 5^(x-3)[124] = (124)(5^y)
Divide both sides by 124 to get: 5^(x-3) = 5^y
So, we can conclude that 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]

Cheers,
Brent

melanie.espeland wrote:How does 5^x factored by 5^(x-3) become 5^3?

A good thing about factoring is that you can test whether you correctly factored by EXPANDING.
So, for example, if we take 3x + 6 and factor it to get 3(x + 2), we can always check our factoring by expanding.
3(x + 2) = 3x + 6 ...perfect.

Now let's test whether or not 5^x - 5^(x-3) = 5^(x-3)[5^3 - 1]
We'll expand 5^(x-3)[5^3 - 1]

First, we have 5^(x-3)[5^3]
Since the bases are the same, we ADD the exponents.
(x-3) + 3 = x
Great, so 5^(x-3)[5^3] =5^x

Second, we have 5^(x-3)[-1]
We get 5^(x-3)
Great!

So, 5^x - 5^(x-3) = 5^(x-3)[5^3 - 1]

melanie.espeland wrote:Hi Brent,

Also, how are you able to divide each side by 124 when it's only on right side of equation?

We have 124 on BOTH sides.

We got to: 5^(x-3)[5^3 - 1] = (124)(5^y)
Simplify the left side: 5^(x-3)[124] = (124)(5^y)
Divide both sides by 124 to get: 5^(x-3) = 5^y

I hope that helps .

Cheers,
Brent

melanie.espeland wrote: So, just to be clear, when you factor 5^(x-3) out of 5^x, you get 5^3 because
a) the x is pulled out and
b) you need to add three in order to take out -3?

Let's think of it this way:
Say we want to factor x^5 - x^3
Of the two exponents (5 and 3), 3 is smaller.
So, we'll factor out x^3.
So we get: x^5 - x^3 = x^3(something - something else)
At this point, we need to determine the two missing pieces.

First, we need to complete this part: x^3(something) = x^5
Since we're multiplying powers, we need to ask "what must we multiply x^3 by to get x^5?"
The answer is x^2.
In other words, (x^3)(x^2) = x^5
NOTICE that 3 + 2 = 5 [we're adding the exponents here]


The original question is no different.
We want: 5^x = 5^(x-3)(something )
Here, the missing something is 5^3, since we know that [5^(x-3)][5^3] = 5^x

A few more examples:
[2^(x-4)][2^4] = 2^x
[7^(x-1)][7^1] = 7^x
[6^(x-11)][6^11] = 6^x


Does that help?

Cheers,
Brent