If 2^(8x) = 640000, then what is the value of 2^(2x−2)?

\(2^{(8x)} = 640000 = 2^6*10^4\)
\(2^{(2x)} = 2^{6/4}*10^{4/4}\)
\(2^{(2x)} = 2^{3/2}*10\)
\(2^{(2x)} = 2\sqrt{2}*10 = 20\sqrt{2}\)

\(2^{(2x-2)} = 20\sqrt{2}/4 = 5\sqrt{2}\)

Hence Option A is correct
Hit Kudos if you liked it

Another approach:

Given: 2^(8x) = 640000
Raise both sides to the power 1/2 (aka take square root of both sides) to get: [2^(8x)]^(1/2) = 640000^(1/2)
Simplify to get: 2^(4x) = 800
Raise both sides to the power 1/2 (again) to get: [2^(4x)]^(1/2) = 800^(1/2)
Simplify: 2^2x = √800
Simplify: 2^2x = 20√2
Divide both sides by 2² (aka 4) to get: (2^2x)/2² = (20√2)/4
Simplify both sides to get: 2^(2x - 2) = 5√2

Answer: A

RELATED VIDEO

In the step

2^(2x)=20√2 is OK

Now
2^(2x−2)=2^(2X) / 2^2 = 20√2 / 4 = 5√2

what was your approach to this Bunuel because I'm not quite following the posted solution. Thanks!

Step 1:

\(2^{(8x)} = 640000\);

\((2^{2x})^4 = 2^{10}*5^4\);

\(2^{2x} = 2^{(\frac{5}{2})}*5\);

\(2^{2x} = \sqrt{2^5}*5\);

\(2^{2x} = 20\sqrt{2}\).


Step 2:

\(2^{(2x−2)}=\frac{2^{(2x)}}{2^2}\).



Step 2:

Substitute \(2^{2x} = 20\sqrt{2}\) into \(\frac{2^{(2x)}}{2^2}\).:

\(\frac{20\sqrt{2}}{2^2}=5\sqrt{2}\).


Answer: A.

Hope it's clear.
_________________

Similar questions to practice:
https://gmatclub.com/forum/if-3-6x-8-10 ... 98777.html
https://gmatclub.com/forum/if-3-6x-810- ... 54347.html

8. Exponents and Roots of Numbers

• Theory
  Exponents and Roots on the GMAT: Tips and hints
  Cyclicity and the remainders on the GMAT
  
  • Questions
    Units digits, exponents, remainders problems
    Tough and tricky DS exponents and roots questions with detailed solutions
    Tough and tricky PS exponents and roots questions with detailed solutions
    Exponents DS Questions
    Exponents PS Questions
    Roots DS Questions
    Roots PS Questions

Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________

Another approach, although most of these approaches is solving the equation..

Let \(2^{(2x−2)}=y\), so \(y^4=2^{(2x−2)*4}=2^{8x-8}=\frac{2^{8x}}{2^8}=\frac{640000}{2^6*2^2}=\frac{10000}{2^2}\)
so \(y^4=\frac{10000}{2^2}=\frac{10^4}{2^2}....y=\frac{10}{\sqrt{2}}=5\sqrt{2}\)

A
_________________

We need to determine the value of (2^(2x))/2^2 = (2^(2x))/4

If we take the square root of the given equation, we are left with:

2^(4x) = 800

If we take the square root of 2^(4x) = 800, we have:

2^(2x) = √800

2^(2x) = √800 = √400 x √2 = 20√2

Thus, (2^(2x))/4 = 20√2/4 = 5√2

Answer: A
_________________

While the solutions posted above are all elegant, I am sharing here the way to solve backwards from answer choices. This will be helpful in case one is not able to think of the direct approach.

1. \(2^{(2x−2)}\) can be re-written as \(2^{2x}/2^2\) (----> 1)
2. Similarly, \(2^{(8x)} = 640,000\) can be re-written as \(2^{2x(4)}\) \(=\) \(640,000\) to make it inline with the numerator i.e. \(2^{2x}\) from point 1 (----> 2)

Let's start with option A:
\(2^{(2x−2)}\) = \(2^{2x}/2^2\) = \(5\sqrt{2}\)
\(2^{2x}\) \(=\) \(5\sqrt{2} * 2^2\) = \(20\sqrt{2}\)

Plugging the value of \(2^{2x}\) into \(2^{2x(4)}\) from (----> 2)
\((20\sqrt{2})^4\) \(=\) \(640,000\)

As we get the answer from the first option itself, we do not need to test the other answer options

Ans. A

in such questions keep an eye on what is being asked.
here: \(2^{(2x−2)}\)

given: \(2^{(8x)} = 640000\) => \(2^{8x}=2^{10}*5^{4}\)

First, we need 2x in the power, so:
\(2^{2x}=2^{5/2}*5\)
now we need 2x-2
\(2^{2x-2}=2^{1/2}*5\)

ans: A. 5√2

\(\frac{2^{(8x)}}{2^8} = \frac{640000}{2^8}\)
\(2^{(8x-8)} = \frac{2^6*2^4*5^4}{2^8}\)
\(2^{(8x-8)} = 2^2*5^4\)
\(2^{(8x-8)}^{\frac{1}{4}} = (2^2*5^4)^{\frac{1}{4}}\)
\(2^{(2x-2)} = \sqrt{2}*5\)


_________________

2^8x = 640000 = 2^6 * 10^4

2^2x = 2^(8x/4) = 2^(6/4) * 10 = 2^(3/2) * 10

now , 2^(2x-2) = 2^2x / 2 = 2^ (3/2) * 10 / 2 = 2^(1.5-2) / 10 = 10/√2 = 5√2

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________