# An Application of Change of Base

Today we're going to take a look at a strategy for logarithm multiplication. First, let's take a look at a problem.

(Note: all variables in this post are assumed to be in their respective domains)

What is $\log_227\times\log_364?$ One of the first things you should notice about this is that $27$ and $64$ are powers of the bases of the logarithms, but they are paired with the wrong base. However, you might be tempted to say that $\log_227\times\log_364=\log_264\times\log_327=6\times3=18,$ but you would first need to prove that this is true: $\log_mA\times\log_nB=\log_nA\times\log_mB.$ $\text{...........................................................................................................}$ Let's take a look at what change of base is. Change of base is a technique often used in approximating values of logarithms. Most scientific calculators cannot find logarithms with bases that aren't $2,$ $10,$ or $e,$ but all of them have at least a button that finds the natural logarithm of a number. Change of base lets you split a logarithm so it is a lot easier to find on a calculator. Here's the formula. $\log_ab=\dfrac{\log_kb}{\log_ka}$ If you take the logarithm base $a$ of a number $b,$ then you can take the logarithm base $k$ of $a$ and divide it by the logarithm base $k$ of $b$ to find the logarithm base $a$ of $b.$ For example, if you are told to find $\log_37,$ using only a scientific calculator, then change of base allows you to do this. $\log_37=\dfrac{\ln7}{\ln3}\approx\dfrac{1.945910}{1.098612}\approx1.771244$ (Note: approximation signs are used because of rounding to $6$ decimal places.) $\text{...........................................................................................................}$ So now let's take a look back at the generalization. We are trying to prove that $\log_mA\times\log_nB=\log_nA\times\log_mB.$ Use change of base to split the logarithms. $\log_mA\times\log_nB=\dfrac{\ln A}{\ln m}\times\dfrac{\ln B}{\ln n}$ Using the communative property of multiplication, you get this. $\dfrac{\ln A}{\ln m}\times\dfrac{\ln B}{\ln n}=\dfrac{\ln A}{\ln n}\times\dfrac{\ln B}{\ln m}$ Finally, using reverse change of base, you can do this. $\dfrac{\ln A}{\ln n}\times\dfrac{\ln B}{\ln m}=\log_nA\times\log_mB$ So that's it! We have proved that $\log_mA\times\log_nB=\log_nA\times\log_mB.$ Now we can go back to the original problem.

To find $\log_227\times\log_364,$ you can switch the bases and say that $\log_227\times\log_364=\log_327\times\log_264=3\times6=\boxed{18}.$ $\text{...........................................................................................................}$ This idea can also be applied over multiple multiplications too. Take a look at this problem.

$\text{Find }\prod_{k=2}^{2047}\log_k(k+1)$ This is equal to $\log_23\times\log_34\times\log_45\times\ldots\times\log_{2047}2048.$ Using the base-switching strategy, you can rearrange this to $\log_22048\times\log_33\times\log_44\times\ldots\times\log_{2047}2047.$ Everything to the right of $\log_22048$ is equal to $1,$ so the value is equal to $\log_22048=\boxed{11}.$ $\text{...........................................................................................................}$ $\textbf{Problems}$

$\boxed{1}$ Prove that $\log_1x$ does not exist.

$\boxed{2}$ Prove that $\displaystyle\prod_{k=1}^\infty\log_{2k}(2k+3)$ is divergent.

$\boxed{3}$ Find the value of $\displaystyle\lim_{n\rightarrow1}\left(\log_nk\times\log_hn\right).$ $\text{...........................................................................................................}$ So you learned about multiplying logarithms. I hope that this strategy helps you in the future. Check the tag #TrevorsTips occasionally to see more problem solving strategies from me. Thanks for reading this post!

Note by Trevor B.
6 years, 7 months ago

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I'm not that good with series, but I'll try: It's obvious that $\log_{2k} (2k+3) > \log_{2k} (2k+2)$ for every $k \ge 1$; so, if $\displaystyle \prod_{k=1}^\infty \log_{2k} (2k+2)$ diverges, then $\displaystyle \prod_{k=1}^\infty \log_{2k} (2k+3)$ also diverges. But the former is equal to $\lim_{k \to +\infty} \dfrac{\log 4}{\log 2} \cdot \dfrac{\log 6}{\log 4} \cdot \dfrac{\log 8}{\log 6} \cdots \dfrac{\log (2k+2)}{\log 2k} =\lim_{k \to +\infty} \dfrac{\log (2k+2)}{\log 2} = \infty$ So, the series converges.

As for the third, we have, for $n \ne 1, \log_n k \cdot \log_h n = \dfrac{\log k}{\log n} \cdot \dfrac{\log n}{\log h} = \dfrac{\log k}{\log h} = \log_h\ k$; so, the limit is $\log_h k$.

- 6 years, 7 months ago

Perfect! That is exactly how I intended the second question to be solved!

Do you know how to justify that $\displaystyle\lim_{n\rightarrow1}\log_nn=1?$ There's a couple different methods.

- 6 years, 7 months ago

It's simple: $\log_n n = \dfrac{\log n}{\log n}$. Since $n \ne 1$ - by the limit, it goes to 1 - we have $\log n \ne 0$ and we can cancel them out, staying with an 1.

- 6 years, 7 months ago

Very good. Just out of curiosity, are you aware of l'Hospital's Rule, which is another technique that can be used here?

It works really well here because you don't have to evaluate the function at values close to $1.$ Granted, it is a really simple function, but if you have other functions that are difficult to evaluate and you have an indeterminate form, you can use it.

Here, you do this. $\lim_{n\rightarrow1}\log_nn=\dfrac{\ln n}{\ln n}=\lim_{n\rightarrow1}\dfrac{\frac{1}{n}}{\frac{1}{n}}=\dfrac{1}{1}=\boxed{1}$

- 6 years, 7 months ago

What if the second question was on $\prod \log_{2k} (2k+1)$? Then the telescoping trick used by Joao would not work immediately

The question I want to get to is:
What is the largest value of $\alpha$ such that $\prod \log_{2k} (2k+ \alpha)$ is finite?

Note that $\alpha = 0$ works. Is that the largest possible value?

Staff - 6 years, 7 months ago

@Trevor B. Can you add this to the Number Bases Wiki, likely the Converting to Different Bases page.

Staff - 5 years, 11 months ago

As far as the first problem goes, it's impossible because if $x$ isn't one, then it's impossible to multiply one's and get anything but one. And if it is one, they one to the anything is 1!

- 6 years, 7 months ago

It's possible to prove mathematically that it doesn't exist with a strategy in this discussion. Can you provide a proof based off of the information in the discussion?

- 6 years, 7 months ago

Using the change of base, we get $\log_1 x = \dfrac{\ln x}{\ln 1}$. But $\ln 1 = 0$ and we can't have division by zero.

- 6 years, 7 months ago

Yup. But it doesn't have to be ln. It could be log, or log base anything, as long as it's consistent for both logarithms.

- 6 years, 7 months ago

I know; I would generalize to any base, but I didn't think it was necessary.

- 6 years, 7 months ago

Ok, just checking.

- 6 years, 7 months ago

Correct! Can you try the other two?

- 6 years, 7 months ago

Happy birthday Trevor :).

- 6 years, 7 months ago

Thanks!

- 6 years, 7 months ago

what is meant by divergence of a logarithm

- 6 years, 2 months ago