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I've been using the following method to derive/remember the logarithm base conversion formula: If I want to convert $\log_a(x)$ to an expression in base $b$, I say, $$a^{\log_a(x)}=x\\ \log_b(a^{\log_a(x)})=\log_b(x)\\ \log_a(x)\log_b(a)=\log_b(x)\\ \log_a(x)=\frac{\log_b(x)}{\log_b(a)}$$ It kind of feels like I'm working backwards, and I was wondering if there's a more direct way to go about it. I tried: $$\log_a(x)=\log_a(b^{\log_b(x)})\\ \log_a(x)=\log_b(x)\log_a(b)$$ but couldn't rid myself of the $\log_a$ in the right-hand side of the equation. It occurred to me that if I could rewrite the "$a$" subscript as "$b^{\log_b(a)}$", I might be onto something. (Or might not.) Does the notation ever get used like that, where you perform a substitution in a subscript?

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Deriving and remembering are different things. If all you want is to remember, do remember this: All logarithm functions are proportional to each other. Thus $$\log_a(x)=C\log_b(x)$$ for some constant $C$. To find out the value of $C$, insert $x=b$ and remember that $\log_b(b)=1$.

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It is not common, but substitution is universal. Anywhere there is something that represents a number, you may put there anything else that represents the same number. This is at the very heart of mathematics, the fact that numbers transcend any representation there of. Oh, and I have no idea how that would help, unless you know how to extract powers from a base. (If you do, show me your powers of magic in the comments. I have been interested in logarithms recently.)

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Look at the two equations you have produced:

$$\log_a(x) = \frac{\log_b(x)}{\log_b(a)} = \log_b(x) \cdot \log_a(b) , $$

all of which is correct. This suggests a couple of things. One is the relation

$$\log_a(b) = \frac{1}{\log_b(a)} ,$$

so you have found one way to manipulate logarithmic values.

You also have two equivalent expressions for the "change-of-base" formula; both are taught in various texts -- I'm not sure which is more commonly used. But this gives you a couple ways to remember the formula:

$$ \log_{\text{new base}} (x) = \frac{\log_{\text{old base}}(x)}{\log_\text{old base}(\text{new base})}$$

or

$$ \log_\text{new base}(\text{old base}) \cdot \log_{\text{old base}} (x) = \log_{\text{new base}}(x) ,$$

[this second one was obtained by relabeling, and not just by algebra]

which sort of looks like a "cancellation rule" (but we shouldn't call it that)...

Here's an example:

$$\frac{\log_e (35)}{\log_e (10)} = \log_{10} (35) \Rightarrow \frac{3.55535}{2.302585} \approx 1.544068... \approx \log_{10}(35) $$

You can also use the relation you found to find $\log_{10} (e) = \frac{1}{\log_e (10)} \approx \frac{1}{2.302585} \approx 0.434294 $ . A calculator actually only computes natural logarithms $\log_e (x)$ or $\ln (x)$ , and uses the change-of-base formula (with the value of $\ln 10$ "hard-wired in") to calculate and display common logarithms $\log_{10} (x)$ . If you need logarithms to a base other than $e$ or 10 , for which the calculator has no buttons, you definitely need the "change-of-base formula" (though I've seen some recent calculators that have a "$\log_{\Box} x$" button)...

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