Question

If \[ 2^a=4^b=8^c \] and \[ abc=288, \] then \[ \frac{1}{2a}+\frac{1}{4b}+\frac{1}{8c}= \]

• A. \(\frac{9}{74}\)
• B. \(\frac{11}{86}\)
• C. \(\frac{13}{82}\)
• D. \(\frac{11}{72}\)

Hint:

Convert all powers to the same base first. This makes the relation between exponents straightforward.

Answer 1

The Correct Option is D

Concept: When exponential expressions are equal, equate them to a common variable.

Step 1: Assume common value.
Let: \[ 2^a=4^b=8^c=k \] Convert into base \(2\): \[ 2^a=(2^2)^b=(2^3)^c \] \[ 2^a=2^{2b}=2^{3c} \] Thus: \[ a=2b=3c \]

Step 2: Express in one variable.
Let: \[ a=t \] Then: \[ b=\frac t2,\quad c=\frac t3 \] Given: \[ abc=288 \] Substitute: \[ t\cdot \frac t2 \cdot \frac t3=288 \] \[ \frac{t^3}{6}=288 \] \[ t^3=1728 \] \[ t=12 \] So: \[ a=12,\;b=6,\;c=4 \]

Step 3: Find the required sum.
\[ \frac1{2a}+\frac1{4b}+\frac1{8c} \] Substitute: \[ =\frac1{24}+\frac1{24}+\frac1{32} \] Take LCM \(=96\): \[ =\frac4{96}+\frac4{96}+\frac3{96} \] \[ =\frac{11}{96} \] Thus, the required answer is: \[ \boxed{\frac{11}{96}} \]