# Thread: Logarithms..

1. ## Logarithms..

Hello

I need some help solving this logarithm

42 = e^(-1.7+b) + 3.2

I know that the first step should be log.e^(-1.7+b) and then (-1.7+b)log.e but what happens later? and what happens with 3.2 and 42?

Thank you (:

2. Originally Posted by Fifi
Hello

I need some help solving this logarithm

42 = e^(-1.7+b) + 3.2

I know that the first step should be log.e^(-1.7+b) and then (-1.7+b)log.e but what happens later? and what happens with 3.2 and 42?

Thank you (:
Rearrange the equation so the exponential is on one side and everying else on the other, then take logs.

That is:

42 -3.2 = e^(-1.7+b)

CB

3. Originally Posted by CaptainBlack
Rearrange the equation so the exponential is on one side and everying else on the other, then take logs.

That is:

42 -3.2 = e^(-1.7+b)

CB
Like this?

log.39.8 = (-1.7+b)log.e
log.39.8/log.e = -1.7+b

Edit: Uhm.. it doesn't work out on the calculator with having e alone..?

4. Originally Posted by Fifi
Like this?

log.39.8 = (-1.7+b)log.e
log.39.8/log.e = -1.7+b

Edit: Uhm.. it doesn't work out on the calculator with having e alone..?
Are these logs log to the base 10? Then $\log_{10} 39.8 = (-1.7+b)\log_{10} e \Rightarrow \frac{\log_{10} 39.8}{\log_{10} e} = -1.7 + b$.

But if you're using a scientific calculator then you should have a log to base e button, in which case you can say $\log_e 39.8 = -1.7+b$.

5. Originally Posted by mr fantastic
Are these logs log to the base 10? Then $\log_{10} 39.8 = (-1.7+b)\log_{10} e \Rightarrow \frac{\log_{10} 39.8}{\log_{10} e} = -1.7 + b$.

But if you're using a scientific calculator then you should have a log to base e button, in which case you can say $\log_e 39.8 = -1.7+b$.
They are not to the base of 10, from what I know at least. I am sorry if I confused you with the dot. Thank you for the help though! And oh, I can't find the log to the base of e button on my calculator, I'm using TI-84 (texas). Sorry for asking a lot.

6. Originally Posted by Fifi
They are not to the base of 10, from what I know at least. I am sorry if I confused you with the dot. Thank you for the help though! And oh, I can't find the log to the base of e button on my calculator, I'm using TI-84 (texas). Sorry for asking a lot.
it will be the button that has ln written on it (or above it).

7. Originally Posted by mr fantastic
it will be the button that has ln written on it (or above it).
Yes I know where the Log button is.. hmm.
Weird, I don't seem to get the correct result actually... should it be log(e^38.8)? because e is demanding something to the power of... and I suppose you mean that e should be within the brackets. There's no log(to the base of e) button.. if that is what I should have...

8. Originally Posted by Fifi
Yes I know where the Log button is.. hmm.
Weird, I don't seem to get the correct result actually... should it be log(e^38.8)? because e is demanding something to the power of... and I suppose you mean that e should be within the brackets. There's no log(to the base of e) button.. if that is what I should have...
I have just told you how to find the log base e button. Irregardless of this, you don't need it .... Just use the log button (which is actually the log base 10 button).

I don't know why you would be wanting to use the 'e to the power of' button .... What I have posted does not involve exponentials.

You need to read again the replies you've been given and then press the correct buttons on your calculator.

9. Originally Posted by mr fantastic
I have just told you how to find the log base e button. Irregardless of this, you don't need it .... Just use the log button (which is actually the log base 10 button).

I don't know why you would be wanting to use the 'e to the power of' button .... What I have posted does not involve exponentials.

You need to read again the replies you've been given and then press the correct buttons on your calculator.
Found this on the net about bases by the way, if someone else needs it: ln(2) is just the log of 2, and ln(2)/ln(10) is log of 2, base 10. You can also see that ln(2)/ln(e), the base e log which is the natural log, will be just ln(2) since ln(e) evaluates to 1.

The thing is, I have to use the e button and.. well, it is apparently only e to the power of something on my calculator, should I just disregard e? What should b be? I get 3.70

I am a bit confused and it could be due to my rather poor English...

10. Originally Posted by Fifi
Found this on the net about bases by the way, if someone else needs it: ln(2) is just the log of 2, and ln(2)/ln(10) is log of 2, base 10. You can also see that ln(2)/ln(e), the base e log which is the natural log, will be just ln(2) since ln(e) evaluates to 1.

The thing is, I have to use the e button and.. well, it is apparently only e to the power of something on my calculator, should I just disregard e? What should b be? I get 3.70

I am a bit confused and it could be due to my rather poor English...
Does the 'e button' have writing above it? What color is the writing? The real problem is that you're not familiar with how to use your calculator. I suggest you go to the user manual or get help from your teacher.