proving that log(u/v).log(s/t) equals log(u/s).log(v/t) + log(u/t).log(s/v)

[YAAMBE]

If u, v, s, t are all positive show that

log(u/v).log(s/t) = log(u/s).log(v/t) + log(u/t).log(s/v)

the logarithms all being to the same base.

How would I even start with this one?

 

[stapel]

If u, v, s, t are all positive show that

log(u/v).log(s/t) = log(u/s).log(v/t) + log(u/t).log(s/v)

the logarithms all being to the same base.

How would I even start with this one?

I will guess that the decimal point "." is being used here to indicate multiplication "*". (For notes on standard web-safe text-only formatting, try here.) So the left-hand side (LHS) is:

. . . . .[imath]\log\left(\dfrac{u}{v}\right)\, \cdot\, \log\left(\dfrac{s}{t}\right)[/imath]

...and the right-hand side (RHS) is:

. . . . .[imath]\log\left(\dfrac{u}{s}\right)\, \cdot\, \log\left(\dfrac{v}{t}\right)\, +\, \log\left(\dfrac{u}{t}\right)\, \cdot\, \log\left(\dfrac{s}{v}\right)[/imath]

I would probably start with the rules for logs, expanding each side (on scratch paper). Then I'd fiddle around and see if I can stumble upon anything useful, keeping the "trick" of the change-of-base formula in the back of my head. So, expanding using logs rules (here), I get:

. . . . .[imath]{LHS: }\, \left(\log(u)\, -\, \log(v)\right)\, \cdot\, \left(\log(s)\, -\, \log(t)\right)[/imath]

. . . . .[imath]{LHS: }\, \log(u)\, \log(s)\, -\, \log(v)\, \log(s)\, -\, \log(u)\, \log(t)\, +\, \log(v)\, \log(t)[/imath]

. . . . .[imath]{RHS: }\, \left(\log(u)\, -\, \log(s)\right)\, \cdot\, \left(\log(v)\, -\, \log(t)\right)\, +\, \left(\log(u)\, -\, \log(t)\right)\, \cdot\, \left(\log(s)\, -\, \log(v)\right)[/imath]

. . . . .[imath]{RHS: }\, \bigg(\, \log(u)\, \log(v)\, -\, \log(s)\, \log(v)\, -\, \log(u)\, \log(t)\, +\, \log(s)\, \log(t)\, \bigg)\, [/imath]

. . . . . . . . . . . . .[imath]+\, \bigg(\, \log(u)\, \log(s)\, -\, \log(t)\, \log(s)\, -\, \log(u)\, \log(v)\, +\, \log(t)\, \log(v)\, \bigg)[/imath]

. . . . .[imath]{RHS: }\, \log(u)\, \log(v)\, -\, \log(u)\, \log(v)\, -\, \log(s)\, \log(v)\, -\, \log(u)\, \log(t)\, [/imath]

. . . . . . . . . . . . .[imath]+\, \log(s)\, \log(t)\, -\, \log(t)\, \log(s)\, +\, \log(u)\, \log(s)\, +\, \log(t)\, \log(v)[/imath]

Simplify the RHS. What do you get? Can you "see" now how to start on the RHS, work to the "middle", and then work the rest of the way (kind-of cheating, working backwards from your scratch paper) back to the original form of the LHS?

 

[Boss 01]

Can you decode please
. .[imath]\displaystyle \mbox{LHS: }\, \left(\log(u)\, -\, \log(v)\right)\, \cdot\, \left(\log(s)\, -\, \log(t)\right)[/imath]

. . . . .[imath]\displaystyle \mbox{LHS: }\, \log(u)\, \log(s)\, -\, \log(v)\, \log(s)\, -\, \log(u)\, \log(t)\, +\, \log(v)\, \log(t)[/imath]

. . . . .[imath]\displaystyle \mbox{RHS: }\, \left(\log(u)\, -\, \log(s)\right)\, \cdot\, \left(\log(v)\, -\, \log(t)\right)\, +\, \left(\log(u)\, -\, \log(t)\right)\, \cdot\, \left(\log(s)\, -\, \log(v)\right)[/imath]

. . . . .[imath]\displaystyle \mbox{RHS: }\, \bigg(\, \log(u)\, \log(v)\, -\, \log(s)\, \log(v)\, -\, \log(u)\, \log(t)\, +\, \log(s)\, \log(t)\, \bigg)\,[/imath]

 

[topsquark]

Can you decode please
. .[imath]\displaystyle \mbox{LHS: }\, \left(\log(u)\, -\, \log(v)\right)\, \cdot\, \left(\log(s)\, -\, \log(t)\right)[/imath]

. . . . .[imath]\displaystyle \mbox{LHS: }\, \log(u)\, \log(s)\, -\, \log(v)\, \log(s)\, -\, \log(u)\, \log(t)\, +\, \log(v)\, \log(t)[/imath]

. . . . .[imath]\displaystyle \mbox{RHS: }\, \left(\log(u)\, -\, \log(s)\right)\, \cdot\, \left(\log(v)\, -\, \log(t)\right)\, +\, \left(\log(u)\, -\, \log(t)\right)\, \cdot\, \left(\log(s)\, -\, \log(v)\right)[/imath]

. . . . .[imath]\displaystyle \mbox{RHS: }\, \bigg(\, \log(u)\, \log(v)\, -\, \log(s)\, \log(v)\, -\, \log(u)\, \log(t)\, +\, \log(s)\, \log(t)\, \bigg)\,[/imath]

\mbox is not a usable command line with our complier. They look like simple equations, just type them out and don't try to be fancy.

The first one should simply be
( log(u) - log(v) ) ( log(s) -log(t) )
right?

Just code that as
[imath]( log(u) - log(v) ) ( log(s) -log(t) )[/imath]

-Dan