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I happen to use this heavy math for the first time for a long time (if ever) and don't know how to bite it.

Given: $$\begin{align} A &= 1.45\\ B &= 4.1\\ C &= 14\\ \frac1A + \frac1B + \frac1C&=100\%\\\ \end{align} $$ I want to find $X$, $Y$ and $Z$ such that $$ \frac1X + \frac1Y +\frac1Z = 112\% $$ I have $$ X=K^{\log_2A}, Y=K^{\log_2B}, Z=K^{\log_2C} $$ and need to find a suitable value of $K$.

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  • $\begingroup$ What does $\,\log(A,2)\,$ mean? Logarithm of two in base $\,A\,$ or what? $\endgroup$ – DonAntonio Jul 19 '13 at 19:07
  • $\begingroup$ Log(A,2) means the logarithm to the base 2 of A, eg. log(8,2)=3 $\endgroup$ – nutship Jul 19 '13 at 19:08
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You say that $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$ is 100%, but since we know exactly what the value is, we can exactly figure out what quantity 100% refers to:

$$\frac{1}{1.45}+\frac{1}{4.1} + \frac{1}{14} \approx 1.005$$

Therefore,

$$\frac{1}{X}+\frac{1}{Y} + \frac{1}{Z} \approx 1.12\cdot 1.005 = 1.1256$$

Now, we have

$$\frac{1}{k^{\log_2 1.45}}+\frac{1}{k^{\log_2 4.1}} + \frac{1}{k^{\log_2 14}} = 1.1256$$

The exponents are just numbers, so although we probably can't find an analytic solution, we can probably find a numerical solution.

Recall that $\log_b a = \frac{\log a}{\log b}$, so we can compute these numbers:

$$\frac{1}{k^{0.5361}}+\frac{1}{k^{2.0356}}+\frac{1}{k^{3.8074}} = 1.1256$$

Using some numerical software, we find $k \approx 1.81685$.

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  • $\begingroup$ $\frac{1}{1.45}+\frac{1}{4.1} + \frac{1}{14} = 1.005$ -- not exactly. Closer to $1.004986183$. But not quite. $\endgroup$ – tomasz Jul 19 '13 at 19:14
  • $\begingroup$ Sorry, I should put approximate signs there. $\endgroup$ – Emily Jul 19 '13 at 19:14
  • $\begingroup$ Unless otherwise stated, an approximation to the third digit after the decimal period seems enough, imo. $\endgroup$ – DonAntonio Jul 19 '13 at 19:15
  • $\begingroup$ Can't we just assume $\frac{1}{1.45}+\frac{1}{4.1} + \frac{1}{14} = 1.00$ ? $\endgroup$ – nutship Jul 19 '13 at 19:15
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    $\begingroup$ @DonAntonio Probably a transcription error, good catch! ed: yep, I had multiplied by 1.112, not 1.12!! $\endgroup$ – Emily Jul 19 '13 at 19:38
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Hints:

We're given

$$100\%=\frac1A+\frac1B+\frac1C=\frac1{1.45}+\frac1{4.1}+\frac1{14}=1.005\implies$$

$$\frac1{K^{\log_21.45}}+\frac1{K^{\log_24.1}}+\frac1{K^{\log_214}}=1.12\cdot1.005\iff$$

$$\frac1{K^{0.536}}+\frac1{K^{2.036}}+\frac1{K^{3.807}}=1.1256\;\ldots$$

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