## let log 4=.45,log 3=.62 Find log(36) and log (4/3)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
sanman777
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### let log 4=.45,log 3=.62 Find log(36) and log (4/3)

let log 4=.45,log 3=.62 Find log(36) and log (4/3)

need help also

stapel_eliz
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let log 4=.45,log 3=.62 Find log(36) and log (4/3)

need help also
Use log rules to break the log(36) and the log(4/3) into logs only in terms of 3 and 4. For instance:

. . . . .log(12) = log(3*4) = log(3) + log(4) = 0.45 + 0.62 = 1.07

sanman777
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### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)

so log(36)=log(6)+log(6)=.778=.778=1.556 is that right?

stapel_eliz
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so log(36)=log(6)+log(6)=.778=.778=1.556 is that right?
You were given the value of log(6)...? (I'm not seeing that in your original post.)

sanman777
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### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)

log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69
im not sure how to do it

stapel_eliz
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log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69
im not sure how to do it
The way you did it is fine. As the lesson you studied (in the link, provided earlier) explained and demonstrated, you use the log rules to break the given log into separate terms which will relate to the given values. So the missing steps (that you'll probably want to include in your hand-in homework) make the solution look like:

. . . . .log(36) = log(9*4) = log(9) + log(4)

. . . . .= log(32) + log(4) = 2*log(3) + log(4)

. . . . .= (2)(0.62) + 0.45

...and so forth.

The other one works the exact same way.

sanman777
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Joined: Mon Mar 23, 2009 3:28 pm
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### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)

oh ok thanks alot