Question

Can someone help me step-by-step with this Khan Academy Logarithm problem and how to arrive at the answers in the missing fields? (Link is below - I will also paste a screenshot of the particular problem).

https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_basics/e/understanding-logs-as-inverse-exponentials

Answer 1

Here is the problem.

Answer 2

From the first column of table I:
\[\large b^{0.792}=3\]
Right away, you would be able to solve for the unknown \(b\):
\[\large\begin{align*}\log_3b^{0.792}&=\log_33\\
0.792\log_3b&=1\\
\log_3b&=\frac{1}{0.792}\\
\log_3b&=1.262626...\\
3^{\log_3b}&=3^{1.262626...}\\
b&=3^{1.262626...}\\
b&\approx 4.00337
\end{align*}\]

Answer 3

You applied the base 3 to both sides to get base 3 log of 3 because that equals 1!!!!! Holy crap I didn't think of that!

Answer 4

So in the second column of table I, you would fill the open slot with \(b^1\), which is just \(b\).

In the third column, you have to work backwards a bit. You want to find \(x\) such that
\[b^x=7\]
You know what \(b\) is, so you would solve like so:
\[\large\begin{align*}
4.00337^x&=7\\
\log_{4.00337}4.00337^x&=\log_{4.000337}7\\
x\log_{4.00337}4.00337&\approx1.40283\\
x&\approx1.40283\\

\end{align*}\]

Answer 5

I totally get how to get an approximation for b, and how to solve this problem with an intuition of the algebra involved. Thank you so much @SithsAndGiggles.

Answer 6

You're welcome. For table II, I'm not sure yet if you use the same \(b\), but it shouldn't be too difficult to figure it out.