Log5(7)=a Log5(6)=b Log5(4)=c Find log5(4/9)
what's the change of base formula ??
Isn't that when you have a calculator? I can't use calcu Actor but the values give ..
I have to write it using the letters
to find log you have to use change of bae formula \[\huge\rm log_a b = \frac{ \log b }{ \log a }\]
do you have to change that log form to exponential form ?
No. I have to rewrite it with the letters.
ohh with log too ?
then yes just apply change of base formula but don't plug in calculator
I'm sorry. Not find the log, but rewrite sorry!
I don't think that's the correct way.. The answer in my key is 2c-2b
wait a sec i thought those are 4 different questions ?
Oh no. The 4 on top is f the variables I use to rewrite the last log with
I mean 3 on the top
i'm sorry i don't understand what they are asking
@jim_thompson5910
Use the properties of logarithms and the logarithms provided to rewrite each logarithm in terms of the variables given
Based on your log laws " Find log5(4/9) " Just means log5 (4) - log5 (9). So we know log5 (4) = c and log5 (9) = ....
I knew that for c.. But can't find log5(9)
It is like that for all my problems that I'm doing. I can find one but then I can find the other
@jim_thompson5910
OKAY! I have figured something out.. kinda LOL we have log5(4/9) . We want to manipulate the information that we are given to somehow end up with log5(4/9) So Let's firstly think of other fractions that are equivalent to 4/9. ---> The hints are in your given information like log5(6) and log5(4). If we square 4, we get 16 right? Square 6 and we get 36. Now, does 16/36 = 4/9 ? I think yeeeess ^_^ Therefore, we're going to want to use. log5(4/9) =log5(4)^2 - log5(6)^2 (then bring the 2 down to the front of the log) = 2log5(4) - 2log5(6) = 2c - 2b
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