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So recently, I decided to do some fairly basic trigonometry for fun to see what I was able to remember. I decided to give myself a challenge and attempted to solve $27=\frac{1}{6^x}$ for $x$ to $3$ decimal places. While I was able to solve the equation, I'm not sure that I was right with the answer that I achieved.
Here's how I got my answer:
Set up the equation:
\begin{align}
27 = \frac{1}{6^x}
\end{align}
Multiply both sides by $6^x$:
\begin{align}
27(6^x) = \frac{1}{\cancel{6^x}}(\cancel{6^x})\iff 27(6^x) = 1
\end{align}
Divide both sides by $27$:
\begin{align}
\frac{\cancel{27}(6^x)}{\cancel{27}}=\frac{1}{27} & \iff 6^x = \frac{1}{27}
\end{align}
Rewrite the equation in logarithmic form: $a^b=x \implies \log_ax=b$
\begin{align}
\log_6\frac{1}{27}=x \iff \log_627^{-1}=x \iff -\log_627=x
\end{align}
Simplifying the logarithm, we get $x \approx -1.839$.
This is my question:
Was I correct with the solution that I achieved, and if not, how would I achieve the correct answer?
Best Answer
My approach, which avoids using a calculator, capitalizes on the fact that I accidentally happen to know that $~\log_{10}(2) \approx 0.301, ~\log_{10}(3) \approx 0.477.$
Therefore $~3^{(301/477)} \approx 2.$
Therefore,
$$27^{-1} = 3^{-3} = 6^x = 2^x \times 3^x \implies $$
$$3^{\frac{301x}{477}} \times 3^x \approx 3^{-3} \implies $$
$$\frac{301x}{477} + x \approx -3 \implies$$
$$\frac{778x}{477} \approx -3 \implies $$
$$x \approx \frac{-3 \times 477}{778}.$$
From here, it is just a long division problem that can be done on scratch paper.