How do you solve for k in ${(\frac{1}{2})}^{\frac{10}{k}} = 0.8$ $(1/2)^(10/k) = 0.8$ ?

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Answer 1 · 2016-05-26T10:10:42

$k = 31.06$ $k=31.06$

${(\frac{1}{2})}^{\frac{10}{k}} = 0.8 = \frac{4}{5}$ $(1/2)^(10/k)=0.8=4/5$ . Now taking log on both sides

$(\frac{10}{k}) log (\frac{1}{2}) = log (\frac{4}{5})$ $(10/k)log(1/2)=log(4/5)$ or

$(\frac{10}{k}) = \frac{log (\frac{4}{5})}{log (\frac{1}{2})}$ $(10/k)=log(4/5)/log(1/2)$ or

$10 = k \cdot \frac{log (\frac{4}{5})}{log (\frac{1}{2})}$ $10=k*log(4/5)/log(1/2)$ or

$k = \frac{10}{\frac{log (\frac{4}{5})}{log (\frac{1}{2})}}$ $k=10/(log(4/5)/log(1/2))$ or

$k = 10 \cdot \frac{log (\frac{1}{2})}{log (\frac{4}{5})} = 10 \cdot \frac{- 0.3010}{- 0.09691}$ $k=10*log(1/2)/log(4/5)=10*(-0.3010)/(-0.09691)$

or $k = 31.06$ $k=31.06$

How do you solve for k in (12)10k=0.8(1/2)^(10/k) = 0.8?