Solve for a variable in the power when the base are two different values

Question

I would like to solve for $C$

$$7^C \times 2^{n-C-1} \le \frac{2^n}{100}$$

Real questions. The different base is really throwing me off. I got up to

$$7^C 2^{-C} \le \frac{1}{50} $$

Answer 1

$$\begin{align*}7^C\cdot 100 \le 2^{n-n+C+1} &\iff 7^C\cdot 100 \le 2^{C+1}\\ &\iff 7^C\cdot100\le 2^C \cdot 2 \\&\iff \left(\frac{7}{2}\right)^C\le \frac{2}{100} \\&\iff \ln\left(\frac{7}{2}\right)^C \le \ln \frac{2}{100} \iff C \ln \frac{7}{2}\le \ln \frac{2}{100}\end{align*}$$

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After editing your question, that you reached $7^C2^{-C}\le \frac{1}{50}$ then you only need to write the LHS as a fraction (optional actually) and to apply the logarithm on both sides in order to obtain $$\ln \left(\frac{7}{2}\right)^C\le \ln \frac{1}{50} \iff C \ln \frac{7}{2}\le \ln \frac{1}{50} \overset{\ln \frac72>0}\iff C\le \frac{-\ln 50}{\phantom{+}\ln \frac72}=-3.122$$

Answer 2

Take the $\log$ of both sides, noting that $\log(x)$ is an increasing function for $x > 0$. In particular, we have \begin{equation} C \log{7} + (n-C-1)\log{2} \leq n \log{2} - \log{100}. \end{equation} Now, it's straightforward to get an inequality with $C$ isolated.