Question

Evaluate (647)(0.00916) using logarithms.

Answer 1

a. 0.8490
b. 9.8490 - 10
c. 0.7728
d. 9.7728 - 10

Answer 2

well it is none of these for sure

Answer 3

is there a typo here? it just asks you to multiply these numbers together?

Answer 4

Use logarithms to evaluate (647)(0.00916)

Answer 5

charlie check that again

Answer 6

Use logarithms to evaluate (647)(0.00916).
The common logarithm of the answer is... a. b. c. d.

Answer 7

when I take those and times them together I get 5.2652 take the log of this and that should give the answer

Answer 8

what answer do you have?

Answer 9

-0.824

Answer 10

Oh wait, .721?

Answer 11

I got it do like this take the of each number and than times them

Answer 12

Take what?

Answer 13

log of the first one and than take second and take those results and times them

Answer 14

take thelog of 647

Answer 15

what you get?

Answer 16

2.81

Answer 17

oh you want the log of the answer!

Answer 18

now do the same for the second number

Answer 19

write down the first answer

Answer 20

-2.03 for the second?

Answer 21

now times them togehter and take the log again

Answer 22

I got -5.7043 but I got error when log

Answer 23

just add, don't multiply

Answer 24

\(\log(ab)=\log(a)+\log(b)\)

Answer 25

yes that is right opps.

Answer 26

in any case in the modern age you can do it all in one step
http://www.wolframalpha.com/input/?i=log%28+%28647%29%280.00916%29+%29&a=*FunClash.log-_*Log10.Log-

Answer 27

i get .7728

Answer 28

but due to the () I thought it was times not add.

Answer 29

satelitle73 how did come to add when there are () with numbers inside, most time that means times.

Answer 30

@satellite73

Answer 31

charlie he/she is right

Answer 32

Oh yes, I see answer c. matches

Answer 33

sorry about all the extra the ( ) confused me.

Answer 34

when you (5) (120) this means to times them.

Answer 35

that is why I went that way from the start. Sorry again about all the extra work

Answer 36

No, it's fine. Thank you :)

Answer 37

again it's my pleasure to help as always. :)