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It's not my intent really to revive dead threads somehow, but I only recently stumble on thinking about the log in other ways than usual and I think I can provide an alternative, but useful view on it.

For more details, please see references in my blog as well as these other threads here in math.stackexchange this and this.

How to directly estimate any (well, you know...almost any) logarithm in any base?

Answer: In order to estimate $\log_b x$, estimate the number of times, $y$, that $x$ can be repeatedly divided/multiplied by $b$ until you get $1$.

It might be easier/more precise to estimate first $\log_bx^n$ and divide the result by $n$ instead.

Examples:

1) Let's estimate $\log_{57}345197$

For the sake of speeding up the estimate, let's say the base was $60$. Now $60^3\sim 216000$, so the log base 60 would be way closer to $3$ than to $4$. $57$ being smaller than $60$, we estimate it is $\log_{57}345197\sim3.0-3.5$.

The actual value up to two decimals is $3.15$, which is an error of about $1/30$. Not bad.

2) Estimate $\log_{0.57}3$. Let's consider $\log_{0.5}81\sim -(6.0-6.5)\sim - 6.25$. The sign means we are multiplying here (see the links). Hence, we can estimate to be $\log_{0.57}3$ to be somewhat larger (in absolute value) thus $\log{0.57}3\sim -6.5/4\sim -1.63$. The exact value up to two decimals is $1.95$, about $4/20=20\%$ error this time.

With a little more practice, I reckon one could make faster and more accurate estimates.

Notice that this approach really doesn't require you to memorize any log in any base: You just estimate them all over again if needed; the overhead isn't that much.

Those links provide more examples and a discussion of further topics exploring this view on its own (i.e., with only an indirect reference to exponentials).