If {2^a} = {4^b} = {(8)^c} and abc = 288 then the value \frac{1}{{2a}} + \frac{1}{{4b}} + \frac{1}{{8c}} is given by A. 1/8 B. -(1/8) C. 11/96 D. -(11/96)?

Question

Answer

We've 2^a=2^2b=2^3c , abc=288

implies a=2b=3c ,

Now

From a=2b, we have b=a/2 and from a=3c we have c=a/3

Substituting these values in abc=288

a.a/2.a/3=288

Therefore a=12, Hence b=6 and c=4

Substitute these values in the question asked and the answer will be 11/96.

Answer

Solution:

Given {2^a} = {4^b} = {(8)^c} and abc = 288.

Let's expand the given terms as follows:

{2^a} = {(2^2)^b} = {(2^3)^c} = 2^(2b) = 2^(3c) = 2^a

So, 2b = a and 3c = a

Multiplying both equations, we get:

6bc = a^2

Substituting abc = 288, we get:

6bc = 288^2

bc = 48^2

Now, we need to find the value of \frac{1}{{2a}} \frac{1}{{4b}} \frac{1}{{8c}}

Substituting 2b = a and 3c = a, we get:

\frac{1}{{2a}} \frac{1}{{4b}} \frac{1}{{8c}} = \frac{1}{{2(2b)}} \frac{1}{{4b}} \frac{1}{{8(3c)}} = \frac{1}{2^{2b+1} 4b 8^{3c-1}}

Substituting bc = 48^2, we get:

\frac{1}{{2a}} \frac{1}{{4b}} \frac{1}{{8c}} = \frac{1}{2^{2b+1} 4b 8^{3c-1}} = \frac{1}{2^{2(48)+1} 4(48) 8^{3(48)-1}} = \frac{1}{2^{97} 3^{3}}

Simplifying, we get:

\frac{1}{{2a}} \frac{1}{{4b}} \frac{1}{{8c}} = \frac{1}{2^{97} 3^{3}} = \frac{-11}{96} * 2^{-97}

Therefore, the answer is D. -(11/96)

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