# Lesson: Exponent Laws - Part II

## Comment on Exponent Laws - Part II

### I love the fact that you have

I love the fact that you have collated many questions from various forums into one place. This is by far the best collection of questions that keep a student engaged within the GMAT community. Would like to see a lot more questions.

Thanks! I'm adding questions every day.

### Hi Brent.

Hi Brent.

https://gmatclub.com/forum/if-5-13-9-7-3-15-x-what-is-the-value-of-x-219722.html

We can answer this question by keeping track of the 5's only.
Given: (5^13)(9^7)=3(15^x)
Rewrite as: (5^13)(9^7)=3(5^x)(3^x)
So, it must be true that (5^13) = (5^x)
So, x = 13

Good question.
When I get to (5^13)(9^7)=3(5^x)(3^x), I can see that, if I were to write the LEFT side of this equation as the product of 5's and 9's, I'd have THIRTEEN 5's and SEVEN 9's.

Likewise, I were to write the RIGHT side as the product of 5's and 3's, I'd have x 5's and (x+1) 3's.

Since the left side EQUALS the right sides, it must be the case that we have the same number of 5's on each side.
In other words, it must be true that x = 13

Does that help?

Cheers,
Brent

### Hey Brent,

Hey Brent,

I am trying to solve the following question:

If (1/5)^m * (1/4)^18 = 1/(2(10))^35 , then m =

17

18

34

35

36.

I have simplified the right side to 1/2 * 10^(-35) but I cannot figure out how to proceed. I've looked through the videos but can't finde a suitable one. Any solutions and video recommendations on this topic are greatly appreciated.

Cheers,
Kevin

### Thank you Brent, as always

Thank you Brent, as always much appreciated!

### Hello Brent,

Hello Brent,

If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

Given d= 1/(2^3*5^7) i

Multiply by 2^4/ 2^4
d=2^4/(2^3∗5^7)∗2^4=
2^4/ 2^7 ∗ 5^7 =
2^4/ 10^7 = 16 / 10^7 = 0.0000016
Hence D will have two non-zero digits, 16, when expressed as a decimal.

I have a question concerning the solution, why do we ignore the multiplication of 2^4 x 5^7 ?

Thank you

### Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/if-d-1-2-3-5-7-is-expressed-as-a-terminating-...

Cheers,
Brent

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/if-y-k-y-5k-m-what-is-k-248526.html

If y^k/y^5k = y^m, what is k?

(1) m = 4
(2) y = 4

Of course I applied the exponent law. In your solution you said "we must be certain that the base does not equal 0, 1 or -1, in which case the exponent laws fly out the window."

Do I always keep it in mind when in the question variables are presented as a base? When the base is zero then the result would be equal to zero despite the exponent?

Q: Do I always keep it in mind when in the question variables are presented as a base? When the base is zero then the result would be equal to zero despite the exponent?

Yes, that's correct.
For example, if we know that 1^x = 1^y, we can't make any conclusions about whether x = y (since 1^x and 1^y evaluate to be 1 for ALL values of x and y.
Likewise, if we're given the equation w^x = w^y, we can't make any conclusions about whether x = y, since we don't know anything about the value of w.

Cheers,
Brent

### Hi Brent,

Hi Brent,

I have a query with the below question:

Is y^x > x^y?

(1) y > x
(2) x is a positive number

Initially, while I was working this out, I thought the answer would be C - but then realized that I didn't consider a case where x=2 & y=4 which would result in y^x < x^y.

My question is that how can we identify what will be the right set of values to put in place of x and y to explore the different answers we may get for the x^y & y^x?

Since there could be so many different combinations of X & Y, how can we figure out the possibly good numbers to choose given that we have to finish this question in about 2 mins?

Good question. Testing values can definitely take a while, but there are a few guidelines to consider that may help speed up the process (these are covered in the following video on "good" numbers: https://www.gmatprepnow.com/module/gmat-data-sufficiency/video/1102)

If x = 1 and y = 2, then x^y = 1^2 = 1, and y^x = 2^1 = 2
So, the answer to the target question is, "YES, y^x > x^y"

From here, let's test one small number and one large number.
If x = 2 and y = 10, then x^y = 2^10 = some huge number, and y^x = 10^2 = 100
So, the answer to the target question is, "NO, it is not the case that y^x > x^y"

The main idea here, is to test different pairs of numbers.
Some examples include:
- Two positive values
- Two negative values
- One small positive value and one large positive value
- One positive fractional value and one large positive value
- One negative value and one positive value
etc.

### Alright got it, I guess a lot

Alright got it, I guess a lot of this will come from practice. The tricky thing here is to try and not miss out on the combinations that will give two different values for the given question.

### Hi Brent,

Hi Brent,

Could you help me with this question?

If 5^11 × 4^5 = 5 × 10^k, what is the value of k?

I started doing the following:
5^11 x (2^2)^5 = 5 × 10^k
5^11 x 2^10 = 5 x 10^k

And then got stuck! Thanks.

### What's the source of this

I'm happy to help.

Given: 5^11 × 4^5 = 5 × 10^k
Divide both sides by 5 to get: 5^10 × 4^5 = 10^k
Rewrite 4 as 2^2 to get: 5^10 × (2^2)^5 = 10^k
Simplify to get: 5^10 × 2^10 = 10^k
Since both terms on the left side have the same exponent we can rewrite as: (5 x 2)^10 = 10^k
Simplify: 10^10 = 10^k
This means k = 10

Cheers,
Brent

### Brent,

Brent,

My mistake, I typed it in a very confusing format! It is indeed a PS question. The 1 & 2 were just showcasing my attempted steps.

I understand the solution now, thank you for your help!!

### Ahhh, that makes sense :-)

Ahhh, that makes sense :-)
I've edited your original post as well as my reply. (I'll delete these last two comments so as not confuse anyone :-)

Cheers,
Brent