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!