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Re: If 2^(8x) = 640000, then what is the value of 2^(2x−2)?
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13 Jan 2018, 07:01

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Bunuel wrote:

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:

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

If 2^(8x) = 640000, then what is the value of 2^(2x−2)?
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13 Jan 2018, 07:22

Bunuel wrote:

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 \(y^4=\frac{10000}{2^2}=\frac{10^4}{2^2}....y=\frac{10}{\sqrt{2}}=5\sqrt{2}\)