solve for K: log base 4 (5k+4)=3
4(5k+4)=3 20k+16=3 -16 -16 20k=-13/20 check: 4(5(-13/20)+4)=3 k=-13/20
thanks , so whenever you hvae log base ___ (any given #) you just multiply that out?
I do not know :/ sorry I just worked out the problem which like that it gives me the answer
actually, do it like this: log_4(5k+4)=3 5k+4=4^3 5k=4^3-4 k=12
i am confused.... what does the log_4 do?
since you have the log with base 4, if you raise the entire left side of the equation to be an exponent of the base, it will drop out and you will be left with the inside of the log function. For example. 10^log_(10)(x) = x
so the 4 is then placed inside the eqation...?
the 4 is the base of the logarithm. it defines how the logarithm behaves.
so if i had an equation that was like log_6 (eq) =2 then i would make it 6^2 ??
yeah pretty much. you would end up with (eq)=6^2
ok. great thanks :) are you good with logs?
I am. I'm just about to graduate the University of Arizona with a degree in Applied Math. Shoot me an email if you need more help ajkelly@email.arizona.edu
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