Solve for $x$ correct to $3$ decimal places: $27=\frac{1}{6^x}$

Question

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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:

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\begin{align} 27 = \frac{1}{6^x} \end{align}

Multiply both sides by $6^x$:

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\begin{align} 27(6^x) = \frac{1}{\cancel{6^x}}(\cancel{6^x})\iff 27(6^x) = 1 \end{align}

Divide both sides by $27$:

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\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: $a^b=x \implies \log_ax=b$

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\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?

Answer 1

Your solution is correct, but you don't need the various passages you show, before taking the logarithm of both sides of the equation. Therefore I suggest the simpler $$ \log(27)=\log\left(\frac{1}{6^x}\right) \tag1 $$ Which turns into $$ \log(27)=0-x\cdot \log(6) \tag2 $$ And solving for $x$ $$ x=-\frac{\log(27)}{\log(6)}\approx-1.839 \tag3 $$ By the way, the logarithm can be taken with any base: $e, 10, 6, 2$ and the result is still the same.

Alternatively, if you want to do something elegant that simplifies away the powers of three, you could rewrite passage (3) as $$ x=-\frac{\log(3^3)}{\log(3\cdot2)} =-\frac{3\cdot \log(3)}{\log(3)+\log(2)}=-\frac{3}{1+\frac{\log(2)}{\log(3)}}\tag4 $$

Answer 2

As it turns out, I was right, but there is a way I could do this faster:

Setting up the equation (again because why not): $$27=\frac{1}{6^x}$$ $$\implies27^{-1}=\left(\frac{1}{6^x}\right)^{-1}$$ $$\implies\frac{1}{27}=6^x$$ $$\implies\log_6\frac{1}{27}=x$$ $$\implies-\log_627=x$$ $$\implies x\,\text{is approximately}\quad-1.839$$

Answer 3

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.