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PS: Exponents

by me_1234 » Wed Dec 17, 2014 2:29 pm
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


Correct answer: C


I'm wondering how to solve this problem? Thank you.
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by Brent@GMATPrepNow » Wed Dec 17, 2014 2:38 pm
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
---------------------
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
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by GMATGuruNY » Wed Dec 17, 2014 2:45 pm
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
Plug in a value for x and solve for y.

Let x=3.
Then:
5³ - 5� = (124)(5^y)
124 = (124)(5^y)
1 = 5^y
y=0. This is our target.

Now plug x=3 into the answers to see which yield our target value of 0.
Only C works:
x-3 = 3-3 = 0.

The correct answer is C.
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by me_1234 » Wed Dec 17, 2014 3:00 pm
Hi Brent,

Thank you, but I'm still not totally getting it. If you are factoring out 5^(x-3), then how did you get 5^(x-3)(5^3 - 1) . I.e., how does 5^x factored by 5^(x-3) become 5^3?

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

Thank you.
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by Brent@GMATPrepNow » Wed Dec 17, 2014 3:13 pm
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
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by me_1234 » Wed Dec 17, 2014 4:37 pm
Thanks Mitch, plugging in seems to the the simplest form of solving.

Thanks Brent, this is making more sense now.

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?
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by Brent@GMATPrepNow » Wed Dec 17, 2014 4:49 pm
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
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by me_1234 » Thu Dec 18, 2014 9:54 am
@Brent@GMATPrepNow, yes that helps, thank you.
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by Mathsbuddy » Fri Dec 19, 2014 2:57 am
5^x - 5^x-3 = (124)(5^y)
5^x - 5^x-3 = (125-1)(5^y)
5^x - 5^x-3 = (5^3-1)(5^y)
5^x - 5^x-3 = 5^(3+y)-5^y
y = x - 3
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by Matt@VeritasPrep » Mon Dec 22, 2014 1:10 pm
Mathsbuddy's approach is the way I'd go. Here's a more legible read:

5ˣ - 5ˣ�³ = 124 * 5ʸ
5ˣ - 5ˣ�³ = (125 - 1) * 5ʸ
5ˣ - 5ˣ�³ = (5³ - 1) * 5ʸ
5ˣ - 5ˣ�³ = 5ʸ�³ - 5ʸ

At this point, x = y + 3 (or y = x - 3, whichever you prefer), since the two larger terms on each side should be the same.
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