The change of base formula of logs means we can find the logarithm of a number using any base simply by changing the base to one we're more familiar with, i.e. the ones that are available to us on our calculator.

So of course it goes something like this

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Rather simple, I know. BUT this is where I'm getting a little confused, in particular this part:

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After evaluating the logs the resulting numbers are the powers/exponents the base 10 is raised to obtain 16 and 4.

We can, of course, represent these in exponential form, in which case we get:

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Because of the powers of quotients property we could evaluate this as

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Clearly this is wrong as $4^4 = 256$, and we were originally trying to find $log_4(16)$ using the change of base property, which of course is $2$.

My question is: Why is this incorrect and where is my error/misstep in logic?


Here's what wrong:

$\log_{10}(4)=.6020...$ says $10^{.6020...}=4$. Similarly, $\log_{10}(16)=1.204...$ says $10^{1.204...}=16$ That is $10^{1.204...-.6020...}=\frac{16}{4}=4$. Where you're confused is on exactly how to read the logs I think. The change of base formula isn't saying $10^{1.204...-.6020...}=4$ It's saying $4^{\frac{1.204...}{6020...}}=16$ which indeed it is.

Another way to clarify this might be to consider the inverses of the log functions. The inverse of $log_{4}(x)$ is $4^x$. So, if we raise the equation you gave to the 10th power we get


So where exactly did you go wrong? It isn't $10^{1.204...-.6020...}$. What you did is the fraction of the inverses of the logs. That is $\frac{f^{-1}(f(16))}{f^{-1}(f(4))}=\frac{16}{4}$. But that isn't equal to your original equation.

  • $\begingroup$ Ahh yes that makes sense now, thank you! $\endgroup$ – Slecker Oct 21 '18 at 20:12

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