C3: is -ln(10) the same ln(0.1)?

I think it is because I got -ln(10) for the answer to a question and the book at the back says ln(0.1) instead of -ln(10).

The question is e^-x=10 in exavt terms.
Can't see how it can turn into ln(0.1) though?

How I got -ln(10):
e^-x=10
-x=ln(10), so x=-ln(10)

Or it could be done like this:
e^-x=1/(e^x)
so 1/e^x=10
multiplying both sides by ln gives 1/x=10/ln, x=ln(10)?

Reply 1

Illiberal Liberal

2

You could have used a calculator to check, but yes it is.

Reply 2

ghostwalker

17

Original post by JacobAlevels

I think it is because I got -ln(10) for the answer to a question and the book at the back says ln(0.1) instead of -ln(10).

The question is e^-x=10 in exavt terms.
Can't see how it can turn into ln(0.1) though?

-ln(10) = 0 - ln(10) = ln(1) - ln(10) = ln (1/10) = ln(0.1)

Or it could be done like this:
e^-x=1/(e^x)
so 1/e^x=10
multiplying both sides by ln gives 1/x=10/ln, x=ln(10)?

The bit in red is gibberish. You can't multiply by "ln". You can take the "ln" of both sides however to get:

ln(1/e^x) = ln(10)

(edited 10 years ago)

Reply 3

Hasufel

16

same answer different form by log laws:

ln(\frac{1}{10})=ln(1)-ln(10)=0-ln(10)

(<-what he said)

Reply 4

Solivagant

18

Original post by JacobAlevels

I think it is because I got -ln(10) for the answer to a question and the book at the back says ln(0.1) instead of -ln(10).

The question is e^-x=10 in exavt terms.
Can't see how it can turn into ln(0.1) though?

How I got -ln(10):
e^-x=10
-x=ln(10), so x=-ln(10)

Or it could be done like this:
e^-x=1/(e^x)
so 1/e^x=10
multiplying both sides by ln gives 1/x=10/ln, x=ln(10)?

Yes, -ln(10=ln(0.1), since aln(b)=ln(b^a)

For your second method, when you have 1/e^x=10, that is equivalent to 1/10=e^x

Reply 5

james22

16

They are the same as ln(x^a)=a*ln(x), but your last 4 lines of working are totally wrong. You don't "multiply" both sides by ln, you take the ln of both sides so you would get ln(1/e^x)=ln(10).