Solving equation with logarithms 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 $K$.
 A: 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 $k \approx 1.81685$.
A: Hints:
We're given
$$100\%=\frac1A+\frac1B+\frac1C=\frac1{1.45}+\frac1{4.1}+\frac1{14}=1.005\implies$$
$$\frac1{K^{\log_21.45}}+\frac1{K^{\log_24.1}}+\frac1{K^{\log_214}}=1.12\cdot1.005\iff$$
$$\frac1{K^{0.536}}+\frac1{K^{2.036}}+\frac1{K^{3.807}}=1.1256\;\ldots$$
