Proving the Change of Base Formula for Logarithms #shorts

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[jonaskoelker]

The way I like to remember the formula is this:

[log base a of b] times [log base b of c] equals [log base a of c]

The pattern matches the transitive property: "[something a b] and [something b c] gives you [something a c]". I think that's why I find it easier to remember: it follows a familiar pattern.

If you divide both sides by [log base a of b] you get the more common

[log base b of c] equals [log base a of c] divided by [log base a of b]

Except with a less common choice of letters.

Also, using my formula, it becomes straightforward to prove:

1. Raise a to the power of both sides. Since this is injective, we have an implication arrow in the direction we need.
2. Simplify "a to the [log base a of c]" to "c"
3. Rewrite "a to the ([log base a of b] times [log base b of c])" to "(a to the [log base a of b]) to the [log base b of c]"
4. Simplify the above to "b to the [log base b of c]"
5. Simplify the above to "c"
6. Now we have "c = c". All the simplifications and rewrites preserved equality, and step 1 was injective, therefore the two expressions we started with must also have been equal.

[nishatmunshi]

Not any k. k bigger than zero and k ≠ 1

[lazaremoanang]

Simple. log(a)b = lnb/lna = lnb×ln10/lna×ln10 = (lnb/ln10)/(lna/ln10) = logb/loga. I don't have the time to watch the video for now.

[anshumanagrawal]

It's like, you make videos just to make mistakes