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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 :)

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