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HI!

 

I have a few problems to solve involving logs. Here's one:

 

I have to write in one log the following:

 

2log base 5 (3) - 3log base 5 (2)

 

So far I went like this:

 

= log5 (3x2) - log5 (2x3)

 

But that would = 0 wouldn't it? You can't express =0 into one log don't you?

 

Also, I found this law: log a (p/q)= (loga (p))- (loga (q))

 

But I don't know how to translate the original equation with that law.

 Aug 1, 2016
 #1
avatar+9665 
0

\(\quad\displaystyle2\log_53 - 3\log_52\\ = \displaystyle\log_5(3^2)-\log_5(2^3)\\ = \displaystyle \log_5(\dfrac{9}{8})\)

\(\boxed {\displaystyle\color{red}a\log b = \log b^a}\\ \color{aqua}and \\\boxed{\displaystyle\color{red}\log a - \log b = \log \dfrac{a}{b}}\)

.
 Aug 1, 2016

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