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 log227×log364?\log_227\times\log_364? One of the first things you should notice about this is that 2727 and 6464 are powers of the bases of the logarithms, but they are paired with the wrong base. However, you might be tempted to say that log227×log364=log264×log327=6×3=18,\log_227\times\log_364=\log_264\times\log_327=6\times3=18, but you would first need to prove that this is true: logmA×lognB=lognA×logmB.\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 2,2, 10,10, or e,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. logab=logkblogka\log_ab=\dfrac{\log_kb}{\log_ka} If you take the logarithm base aa of a number b,b, then you can take the logarithm base kk of aa and divide it by the logarithm base kk of bb to find the logarithm base aa of b.b. For example, if you are told to find log37,\log_37, using only a scientific calculator, then change of base allows you to do this. log37=ln7ln31.9459101.0986121.771244\log_37=\dfrac{\ln7}{\ln3}\approx\dfrac{1.945910}{1.098612}\approx1.771244 (Note: approximation signs are used because of rounding to 66 decimal places.) ...........................................................................................................\text{...........................................................................................................} So now let's take a look back at the generalization. We are trying to prove that logmA×lognB=lognA×logmB.\log_mA\times\log_nB=\log_nA\times\log_mB. Use change of base to split the logarithms. logmA×lognB=lnAlnm×lnBlnn\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. lnAlnm×lnBlnn=lnAlnn×lnBlnm\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. lnAlnn×lnBlnm=lognA×logmB\dfrac{\ln A}{\ln n}\times\dfrac{\ln B}{\ln m}=\log_nA\times\log_mB So that's it! We have proved that logmA×lognB=lognA×logmB.\log_mA\times\log_nB=\log_nA\times\log_mB. Now we can go back to the original problem.

To find log227×log364,\log_227\times\log_364, you can switch the bases and say that log227×log364=log327×log264=3×6=18.\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.

Find k=22047logk(k+1)\text{Find }\prod_{k=2}^{2047}\log_k(k+1) This is equal to log23×log34×log45××log20472048.\log_23\times\log_34\times\log_45\times\ldots\times\log_{2047}2048. Using the base-switching strategy, you can rearrange this to log22048×log33×log44××log20472047.\log_22048\times\log_33\times\log_44\times\ldots\times\log_{2047}2047. Everything to the right of log22048\log_22048 is equal to 1,1, so the value is equal to log22048=11.\log_22048=\boxed{11}. ...........................................................................................................\text{...........................................................................................................} Problems\textbf{Problems}

1\boxed{1} Prove that log1x\log_1x does not exist.

2\boxed{2} Prove that k=1log2k(2k+3)\displaystyle\prod_{k=1}^\infty\log_{2k}(2k+3) is divergent.

3\boxed{3} Find the value of limn1(lognk×loghn).\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.
5 years, 6 months ago

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I'm not that good with series, but I'll try: It's obvious that log2k(2k+3)>log2k(2k+2) \log_{2k} (2k+3) > \log_{2k} (2k+2) for every k1 k \ge 1 ; so, if k=1log2k(2k+2) \displaystyle \prod_{k=1}^\infty \log_{2k} (2k+2) diverges, then k=1log2k(2k+3) \displaystyle \prod_{k=1}^\infty \log_{2k} (2k+3) also diverges. But the former is equal to limk+log4log2log6log4log8log6log(2k+2)log2k=limk+log(2k+2)log2= \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 n1,lognkloghn=logklognlognlogh=logklogh=logh k 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 loghk \log_h k .

João Baptista - 5 years, 6 months ago

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

Do you know how to justify that limn1lognn=1?\displaystyle\lim_{n\rightarrow1}\log_nn=1? There's a couple different methods.

Trevor B. - 5 years, 6 months ago

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It's simple: lognn=lognlogn \log_n n = \dfrac{\log n}{\log n} . Since n1 n \ne 1 - by the limit, it goes to 1 - we have logn0 \log n \ne 0 and we can cancel them out, staying with an 1.

João Baptista - 5 years, 6 months ago

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@João Baptista 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 1.1. 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. limn1lognn=lnnlnn=limn11n1n=11=1\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}

Trevor B. - 5 years, 6 months ago

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What if the second question was on log2k(2k+1) \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 α \alpha such that log2k(2k+α) \prod \log_{2k} (2k+ \alpha) is finite?

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

Calvin Lin Staff - 5 years, 6 months ago

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

Calvin Lin Staff - 4 years, 10 months ago

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As far as the first problem goes, it's impossible because if xx 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!

Finn Hulse - 5 years, 6 months ago

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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?

Trevor B. - 5 years, 6 months ago

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Using the change of base, we get log1x=lnxln1 \log_1 x = \dfrac{\ln x}{\ln 1} . But ln1=0 \ln 1 = 0 and we can't have division by zero.

João Baptista - 5 years, 6 months ago

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@João Baptista 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.

Finn Hulse - 5 years, 6 months ago

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

João Baptista - 5 years, 6 months ago

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@João Baptista Ok, just checking.

Finn Hulse - 5 years, 6 months ago

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@João Baptista Correct! Can you try the other two?

Trevor B. - 5 years, 6 months ago

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Happy birthday Trevor :).

Sam Thompson - 5 years, 6 months ago

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Thanks!

Trevor B. - 5 years, 6 months ago

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what is meant by divergence of a logarithm

Damini damini - 5 years, 1 month ago

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