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by **sanman777** on Mon Mar 23, 2009 4:16 pm

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

need help also

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by **stapel_eliz** on Mon Mar 23, 2009 5:07 pm

sanman777 wrote: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

If you get stuck, please reply showing your progress. Thank you!

by **sanman777** on Mon Mar 23, 2009 5:22 pm

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

by **stapel_eliz** on Mon Mar 23, 2009 5:26 pm

sanman777 wrote: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.) :confused:

by **sanman777** on Mon Mar 23, 2009 5:36 pm

sorry about that
log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69
im not sure how to do it
:confused:

by **stapel_eliz** on Mon Mar 23, 2009 5:40 pm

sanman777 wrote: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. :wink:

The other one works the exact same way.

by **sanman777** on Mon Mar 23, 2009 5:44 pm

oh ok thanks alot :clap:

The Purplemath Forums

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)

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

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

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