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In the step

2^(2x)=20√2 is OK

Now
2^(2x−2)=2^(2X) / 2^2 = 20√2 / 4 = 5√2 :?:
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Bunuel
If \(2^{(8x)} = 640000\), then what is the value of \(2^{(2x−2)}\)?

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

what was your approach to this Bunuel because I'm not quite following the posted solution. Thanks!
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Bunuel
If \(2^{(8x)} = 640000\), then what is the value of \(2^{(2x−2)}\)?

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

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.
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Bunuel
If \(2^{(8x)} = 640000\), then what is the value of \(2^{(2x−2)}\)?

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160


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

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

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

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

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

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

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

A. 5√2
B. 10
C. 10√2
D. 80√2
E. 160

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

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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
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