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

need help also

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

need help also

need help also

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

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

need help also

Use

. . . . .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!

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

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

sorry about that

log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69

im not sure how to do it

log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69

im not sure how to do it

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

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

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

. . . . .= log(3

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

...and so forth.

The other one works the exact same way.

oh ok thanks alot