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## Comment on

Solving Equations with Exponents## Need some help with this one,

## 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

## Brent, I like your method of

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,

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

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

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

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/if-2-4x-3-600-what-is-the-value-of-2-1-x-1733...

Cheers,

Brent

## Hello sir, In one of your

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:/

Question link: https://gmatclub.com/forum/is-5-k-less-than-144719.html#p1996771

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

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

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

## Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/each-of-the-following-equations-has-at-least-...

## 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:/

Question link: https://gmatclub.com/forum/if-r-and-s-are-positive-integers-such-that-2-...

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

Hi Brent can you give your solution for above question

## Here it is: https://gmatclub

Here it is: https://gmatclub.com/forum/each-of-the-following-equations-has-at-least-...