Another question about logarithms

logb + log c = 3

find b in terms of c

I just dont know, I get to log b = 3 - logc

then I seem to get to b = 10^3-logc

then b = 1000 x 1/logc

the answer gets rid of the log and is just 1000/c

why is this?

^ This has been done, I have another question at the bottom :smile:

Please help!

(edited 10 years ago)

Reply 1

maggiehodgson

You know that loga + logb = log(ab)
Can you change 3 into a log?
If you can then you can delog both sides.

Reply 2

joostan

Original post by 3607

The bold's your problem.

\log b = 3 -\log c [br]\Rightarrow \log b = 10^{3-\log c}

Remember that

10^{\log_{10}f(x)} = f(x)

EDIT: The above poster is using my preferred method though this should still work :smile:

(edited 10 years ago)

Reply 3

ghostwalker

Original post by maggiehodgson

If you can then you can delog both sides.

I think the preferred term might be to take the antilog of both sides.

"delog" smacks of deforestation of the jungle and enviromental concerns. +rep! :smile:

Reply 4

maggiehodgson

Thanks.

I'm not very technical am I. What does that +rep mean?

Reply 5

ghostwalker

Original post by maggiehodgson

Thanks.

I'm not very technical am I. What does that +rep mean?

It means I gave your post a "postive reputation". The "1" next to the thumbs up.

My comment was just an amusing aside, not to be taken with any great seriousness.

(edited 10 years ago)

Reply 6

3607

Original post by joostan

The bold's your problem.

\log b = 3 -\log c [br]\Rightarrow \log b = 10^{3-\log c}

Remember that

10^{\log_{10}f(x)} = f(x)

EDIT: The above poster is using my preferred method though this should still work :smile:

Ah, thanks, I don't think I've ever been taught that haha.

Thanks to everyone for trying to help, I found this the most helpful and easy to get :smile:

Reply 7

joostan

Original post by 3607

Ah, thanks, I don't think I've ever been taught that haha.

Thanks to everyone for trying to help, I found this the most helpful and easy to get :smile:

Lol, no prob :smile:

(edited 10 years ago)

Reply 8

maggiehodgson

Original post by ghostwalker

It means I gave your post a "postive reputation". The "1" next to the thumbs up.

My comment was just an amusing aside, not to be taken with any great seriousness.

Thanks for the vote. I'm pretty new to TSR so am not with what can and can't be done.

Also, I didn't really take your comment as a chastisement You've helped me a lot with some mechanics study so I know you're a kind sort of person.

Will be back on mechanics as soon as S1B is out of the way.

Reply 9

TheNote

Original post by joostan

The bold's your problem.

\log b = 3 -\log c [br]\Rightarrow \log b = 10^{3-\log c}

Remember that

10^{\log_{10}f(x)} = f(x)

EDIT: The above poster is using my preferred method though this should still work :smile:

Isn't it b = 10^(3-log c)?
then you cancel the log from the log c because it's log to the base 10 as a power of 10?

(edited 10 years ago)

Reply 10

3607

Original post by joostan

The bold's your problem.

\log b = 3 -\log c [br]\Rightarrow \log b = 10^{3-\log c}

Remember that

10^{\log_{10}f(x)} = f(x)

EDIT: The above poster is using my preferred method though this should still work :smile:

I've got another question for everyone, it's not absolutely vital this is answered, but I would be interested to know why the above is true!

Thanks to anyone who helps :smile:

Reply 11

joostan

Original post by 3607

I've got another question for everyone, it's not absolutely vital this is answered, but I would be interested to know why the above is true!

Thanks to anyone who helps :smile:

The 10 to the power of is the inverse operater of the log. A function and it's inverse acting on something yields that something.
The definition is that