(1/2)^k=√3,(1/3)^m=√2,find value of mk/2

Options

A.1/2,B.1/4,C.1/8,D.2,E.4

We are given the equations:

(12)k=3\left(\frac{1}{2}\right)^k = \sqrt{3}(13)m=2\left(\frac{1}{3}\right)^m = \sqrt{2}

We need to determine the value of mk2\frac{mk}{2}.

Step 1: Express in Logarithmic Form

Taking the natural logarithm (ln) on both sides:

kln⁡(12)=ln⁡(3)k \ln\left(\frac{1}{2}\right) = \ln(\sqrt{3})mln⁡(13)=ln⁡(2)m \ln\left(\frac{1}{3}\right) = \ln(\sqrt{2})

Using logarithm properties:

ln⁡(12)=−ln⁡2,ln⁡(13)=−ln⁡3\ln\left(\frac{1}{2}\right) = -\ln 2, \quad \ln\left(\frac{1}{3}\right) = -\ln 3ln⁡(3)=12ln⁡3,ln⁡(2)=12ln⁡2\ln(\sqrt{3}) = \frac{1}{2} \ln 3, \quad \ln(\sqrt{2}) = \frac{1}{2} \ln 2

Thus, the equations simplify to:

−kln⁡2=12ln⁡3- k \ln 2 = \frac{1}{2} \ln 3−mln⁡3=12ln⁡2- m \ln 3 = \frac{1}{2} \ln 2

Step 2: Solve for kk and mm

Rearrange the equations:

k=−12ln⁡3ln⁡2=12ln⁡3ln⁡2k = -\frac{\frac{1}{2} \ln 3}{\ln 2} = \frac{1}{2} \frac{\ln 3}{\ln 2}m=−12ln⁡2ln⁡3=12ln⁡2ln⁡3m = -\frac{\frac{1}{2} \ln 2}{\ln 3} = \frac{1}{2} \frac{\ln 2}{\ln 3}

Step 3: Compute mk2\frac{mk}{2}

mk2=12×(12ln⁡3ln⁡2×12ln⁡2ln⁡3)\frac{mk}{2} = \frac{1}{2} \times \left(\frac{1}{2} \frac{\ln 3}{\ln 2} \times \frac{1}{2} \frac{\ln 2}{\ln 3} \right)

Since ln⁡3ln⁡2×ln⁡2ln⁡3=1\frac{\ln 3}{\ln 2} \times \frac{\ln 2}{\ln 3} = 1, we get:

mk2=12×14=18\frac{mk}{2} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

Step 4: Select the Correct Answer

Thus, the correct answer is C. 18\frac{1}{8}

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