# Lesson: Solving Equations with Exponents

## Comment on Solving Equations with Exponents

### Need some help with this one,

Need some help with this one, especially how to figure out /solve this part of the equation: [2^(1-x)]^2. Whole question: If 2^4x = 3,600, what is the value of [2^(1-x)]^2 ?

### Hi Robert,

Hi Robert,

We want to EITHER make [2^(1-x)]^2 resemble the given expression, 2^4x OR make the given expression resemble [2^(1-x)]^2

Given: 2^4x = 3,600

Find the square root of both sides (i.e., raise both sides to the power of 1/2)

We get: (2^4x)^1/2 = 3600^1/2

Simplify: 2^2x = 60

Raise both sides to power of -1 to get: (2^2x)^-1 = 60^-1

Simplify: 2^(-2x) = 1/60 [you'll see why I did this shortly]

Our GOAL is to find the value of [2^(1-x)]^2

[2^(1-x)]^2 = 2^(2-2x)

= (2^2)/[2^(-2x)]

= (2^2)/[1/60] since we 2^(-2x) = 1/60

= (2^2)(60)

= 240

### Thanks! However I realized a

Thanks! However I realized a second after I submitted the question that the part I was having difficulty was, was the "Quotient Law" rule that had me perplexed and confused because it was subtracted and not in it's divided form.

### Brent, I like your method of

Brent, I like your method of explanation. With that said how would you explain the solution to the problem below:

Each of the following equations has at least one solution EXCEPT

A. –2^n = (–2)^-n
B. 2^-n = (–2)^n
C. 2^n = (–2)^-n
D. (–2)^n = –2^n
E. (–2)^-n = –2^-n

### When I SCAN the answer

When I SCAN the answer choices, the first thing that comes to mind is that, n = 0 and n = 1 are possible solutions.

So, let's first see what happens when let n = 0
We get:

A. –2^0 = (–2)^-0
NOTE: -2^0 is the same as -(2^0)
So, we have: –(2^0) = (–2)^-0
Evaluate both sides: -1 = 1
Okay, so n = 0 is NOT a solution to the equation.
So, KEEP answer choice A for now

B. 2^-0 = (–2)^0
Evaluate both sides to get: 1 = 1
So, n = 0 IS a solution to the equation.
ELIMINATE B

C. 2^0 = (–2)^-0
Evaluate both sides to get: 1 = 1
So, n = 0 IS a solution to the equation.
ELIMINATE C

D. (–2)^0 = –2^0
Evaluate both sides to get: 1 = -1
So n = 0 is NOT a solution to the equation.
KEEP answer choice D for now

E. (–2)^-0 = –2^-0
Evaluate both sides to get: 1 = -1
So n = 0 is NOT a solution to the equation.
KEEP answer choice E for now

So, we have have answer choices A, D and E remaining

Another possible value for n to consider is 1
So, let's see what happens when let n = 1
We get:

A. –2^1 = (–2)^-1
NOTE: -2^1 is the same as -(2^1)
So, we have: –(2^1) = (–2)^-1
Evaluate both sides: -2 = -1/2
Okay, so n = 1 is NOT a solution to the equation.
So, KEEP answer choice A for now

D. (–2)^1 = –2^1
Evaluate both sides to get: -2 = -2
So, n = 1 IS a solution to the equation.
ELIMINATE D

E. (–2)^-1 = –2^-1
Evaluate both sides to get: -1/2 = -1/2
So, n = 1 IS a solution to the equation.
ELIMINATE E

We're left with the correct answer: A

### Hi Brent,

Hi Brent,

how do we quickly identify which numbers to use? I arbitrarily chose 2 and then 3 and found that all the equations do not equate. Is there a mathematical explanation?

### Since each of the 5 equations

Since each of the 5 equations had SIMILAR expressions on both sides, I chose to test n = 0 and n = 1, since those values would have similar effects on each side of the equation.

If those values didn't yield a single answer, then I'd extend my values to include n = -1 and then perhaps n = 2

I hope that helps.

Cheers,
Brent

### Hi Brent! What happens if the

Hi Brent! What happens if the base is less than 0 (but not -1) ? I mean, If I have, for instance, this equation: (1/10)^(2x)=10^(3x+2). Could I do:10^((-1)(2x))=10^(3x+2) and then solve for -2x=3x+2 ?
Thanks in advance!

### Hi Pau,

Hi Pau,

Your approach is perfect. By rewriting 1/10 as 10^(-1), you are able to then set the exponents equal to each other.

### can you help me with this one

can you help me with this one please:
If 2^(4x) = 3,600, what is the value of [2^(1−x)]² ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

### Hello sir, In one of your

Hello sir, In one of your answer to the above listed question

https://gmatclub.com/forum/is-5-k-less-than-144719.html

When you divided by 5 both sides why didn't you change the inequality sign ?
In one of your lesson of algebra you told to do so ...Please explain i'm a bit confused.

### Question link: https:/

Be careful. When we divide both sides of an inequality by a NEGATIVE value, then we REVERSE the direction of the inequality symbol.

When we divide both sides of an inequality by a POSITIVE value, then the direction of the inequality symbol REMAINS THE SAME.

Here's the video on working with inequalities: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

Cheers,
Brent

Thank you sir :)

### 5^k-5^k-1= 500

5^k - 5^(k-1) = 500
factor out 5^(k-1)
then 5^(k-1)[5 - 1] = 500

Please let me understand this.

### A lot of students struggle

A lot of students struggle with this kind of factoring.

To set things up, consider the following factorizations:
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)
x^8 + x^6 - x^5 = x^5(x^3 + x - 1)

Notice that, each time, the greatest common factor of the terms is the term with the SMALLEST exponent.

So, in the equation 5^k - 5^(k-1) = 500, we have two terms with powers: 5^k and 5^(k-1)
5^(k-1) is the term with the SMALLEST exponent. So we'll factor that out of our expression.
So, we get: 5^k - 5^(k-1) = 5^(k-1)[some terms we need to determine]

First recognize that [5^(k-1)][5^1] = 5^(k-1 + 1) = 5^k
Also recognize that [5^(k-1)][1] = 5^(k-1)

So, it must be the case that 5^k - 5^(k-1) = 5^(k-1)[5 - 1]

Here are a few more examples of this kind of factoring:
w^x + w^(x+5) = w^x(1 + w^5)
2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]

Does that help?

### Thank you.. but still a

Thank you.. but still a slight confusion. I don't understand the digit 5 in 5^(k-1)[5-1]...

w^x + x^(x+5) = w^x(1 + w^5)... Is it w^(x+5) or x^(x+5)?

### We want to show that 5^k - 5^

We want to show that 5^k - 5^(k-1) = 5^(k-1)[5 - 1]
In particular, we want to show that 5^k = 5^(k-1)[5]

First recognize that 5 and 5^1 are the same.
So we could also write: 5^k = [5^(k-1)][5^1]

The Product Law says: (b^j)(b^k) = b^(j+k)
For example, (w^2)(w^4) = w^(2+4) = w^6

Likewise, [5^(k-1)][5^1] = 5^(k-1 + 1) = 5^k

As for your second question (Is it w^(x+5) or x^(x+5)?), you're absolutely right.
It should be: w^x + w^(x+5) = w^x(1 + w^5)
I've edited my earlier response accordingly.

Does that help?

### Can you help me understand

Can you help me understand what we are supposed to figure here in this question: https://gmatclub.com/forum/each-of-the-following-equations-has-at-least-one-solution-except-94119.html

### Hi Brent,

Hi Brent,

For the below question i am stuck at equation r+2s=4 can you help me with this:

https://gmatclub.com/forum/if-r-and-s-are-positive-integers-such-that-2-r-4-s-16-then-2r-211642.html

### Question link: https:/

When r and s are any real numbers, the equation r + 2s = 4 has infinitely many solutions.
However, in this question we're told that r and s are positive integers.
So the only possible solution is r = 2 and s = 1

Does that help?

### https://gmatclub.com/forum

https://gmatclub.com/forum/each-of-the-following-equations-has-at-least-one-solution-except-94119.html

Hi Brent can you give your solution for above question

### Hi Brent, I have a question

Hi Brent, I have a question about using the approximation method in the question: https://gmatclub.com/forum/if-3-k-16-and-2-j-27-then-kj-233302.html

You wrote: ("Given: 3^k = 16
Notice that 3^2 = 9 and 3^3 = 27
Since 16 is approximately halfway between 9 and 27, we can conclude that k is approximately halfway between 2 and 3.
Let's say k ≈ 2.5"). However, wouldn't 16 be closer to 9 because the difference is 7 compared to 27 where the difference is 11?

Also, how do you determine the value of the approximation? Let's say 16 is closer to 9 (aka 3^2) how would you know whether to approximate k to 2.3 or 2.4?

Thank you in advance.

### Question link: https:/

Please note that this approximation strategy is merely a backup approach if we can't solve the question algebraically.
In such cases, we should be happy if we can eliminate some of our options, and then guess from there.

You're correct to say that 16 is closer to 9 than it is to 27.
So we might expect k to have a value around 2.3 or 2.4.
However, since 3^k is not a linear function, our approximations aren't as easy as just a determining the relative position of 16 when compared to 9 and 27.
In fact, if 3^k = 16, then k ≈ 2.524 (which is closer to 3 than it is to 2)