helpHELPHELP LOGS ARE DIFFICULT

Question

niuuu · Answer

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thesmartone · Answer

$\Large log_a (uv) = log_a u + log_a v$

$\Large log_a (\frac{u}{v}) = log_a u - log_a v	$

thesmartone · Answer

Also

$\Large log_a u^n = n ~log_a u$

niuuu · Answer

so would logau^n be the original equation?

thesmartone · Answer

the first term needs that property to simplify it, yes :)

niuuu · Answer

ok I'm brain dead with this one I just have never done well with log, but would I add and subtract the original problem like 2log2^12-log2^4+log2^3= 2log2^5?

thesmartone · Answer

Using this 
Also

$\Large  n ~log_a u = log_a u^n$

change this:

$\Large 2 ~log_2 12 =$

niuuu · Answer

oooh so 2 log2^12= log2^4+log2^3

niuuu · Answer

then would I divide both sides by 2?

thesmartone · Answer

Just change $\Large 2 ~log_2 12 =$ using this formula:

$\Large log_a u^n = n ~log_a u$

niuuu · Answer

log2^12^2?

thesmartone · Answer

yes, and what is 12^2 = ?

niuuu · Answer

isn't that 2 supposed to be there from the front of the log on the other side?

thesmartone · Answer

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I don't think you understand. a is the base of the log. We can't have an exponent before it like 2^12^2.

Let me do an example:

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