stylaria3y

2022-07-21

What log rule was used to simply this expression?

I'm unclear how the left side is equal to the right side.

$365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)=305\mathrm{log}\left(\frac{365}{305}\right)-60$

I know $\mathrm{log}(a)-\mathrm{log}(b)=\mathrm{log}(a/b)$ but if you stick constants before each ln() then how do you apply the rule to get 305 as the constant on the right side of the equation?

I'm unclear how the left side is equal to the right side.

$365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)=305\mathrm{log}\left(\frac{365}{305}\right)-60$

I know $\mathrm{log}(a)-\mathrm{log}(b)=\mathrm{log}(a/b)$ but if you stick constants before each ln() then how do you apply the rule to get 305 as the constant on the right side of the equation?

Bradley Sherman

Beginner2022-07-22Added 17 answers

There are a couple of steps missing.

$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}\text{}365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)\\ & =[365\mathrm{log}(365)-60\mathrm{log}(365)]+[-365+305]-305\mathrm{log}(305)\\ & =305\mathrm{log}(365)-60-305\mathrm{log}(305)\\ & =[305\mathrm{log}(365)-305\mathrm{log}(305)]-60\\ & =305[\mathrm{log}(365)-\mathrm{log}(305)]-60\\ & =305\mathrm{log}(365/305)-60\end{array}$

$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}\text{}365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)\\ & =[365\mathrm{log}(365)-60\mathrm{log}(365)]+[-365+305]-305\mathrm{log}(305)\\ & =305\mathrm{log}(365)-60-305\mathrm{log}(305)\\ & =[305\mathrm{log}(365)-305\mathrm{log}(305)]-60\\ & =305[\mathrm{log}(365)-\mathrm{log}(305)]-60\\ & =305\mathrm{log}(365/305)-60\end{array}$

Makenna Booker

Beginner2022-07-23Added 3 answers

Collect the constants (-365 + 305 = -60), and the terms with $log(365)$

$365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)=305\mathrm{log}(365)-305\mathrm{log}(305)-60$

Now factor out 305, and use the identity you mentioned:

$305(\mathrm{log}(365)-\mathrm{log}(305))-60=305\mathrm{log}(365/305)-60$

$365\mathrm{log}(365)-365-305\mathrm{log}(305)+305-60\mathrm{log}(365)=305\mathrm{log}(365)-305\mathrm{log}(305)-60$

Now factor out 305, and use the identity you mentioned:

$305(\mathrm{log}(365)-\mathrm{log}(305))-60=305\mathrm{log}(365/305)-60$

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