Question

Can you pinpoint a mistake in my steps?

Rewrite the following without logarithms.
\[\ln I = \ln(2V) - [\ln(KR+r) - \ln K + KL]\]

1.
\[2V \div [\ln(KR + r) - \ln K + KL]\]

2.
\[ 2V \div [(KR + r) e^{KL} \div \ln K]\]

3.
\I = 2VK \div (KR + r)e ^{KL} \

Answer 1

What were the steps you took?

Answer 2

\[\ln I = \ln(2V) - [\ln(KR+r) - \ln K + KL]\]
\[2V \div [\ln(KR + r) - \ln(K) + KL]\]
\[ 2V \div [(KR + r) e^{KL} \div \ln K]\]
\[I = 2VK \div (KR + r)e ^{KL}\]

Answer 3

First line is the equation. The 3 other lines are my steps.

Answer 4

\[lnI=\ln(2V)-\ln(K(KR+r))-KL=\ln (\frac{2V}{K(KR+r)})-KL\]

Answer 5

\[lnI=\ln(\frac{2V}{K(KR+r)})-KL\]

Answer 6

Raising e to the power of both sides from there, we get:

\[I = e^{\ln(\frac{2V}{K(KR+r)})-KL}\]
\(=e^{\ln(\frac{2V}{K(KR+r)})}e^{-KL}\)
\(=\frac{2V}{K(KR+r)}e^{-KL}\)

Answer 7

\[e^{lnI}=e^{\ln(\frac{2V}{K(KR+r)})-KL}\]
\[I=e^{\ln (\frac{2V}{K(KR+r)})}e^{-KL}=\frac{2V}{K(KR+r)}e^{-KL}\]

Answer 8

Oh, and we can distribute the denominator a bit better there to look nicer..
\[={2Ve^{-KL} \over K^2R + Kr}\]

Answer 9

\[I=\frac{2V}{e^{KL}(K(KR+r)}\]

Answer 10

Oh, yeah.. Better still.

Answer 11

i forgot a closing parenthesis at the end

Answer 12

\[={2V \over e^{KL}(K^2R + Kr)}\]

Answer 13

Myin, you know you can right click someones expression and view the source to copy/paste right?

Answer 14

Saves having to retype long complicated messes like that when you've already done it once.

Answer 15

i didn't know thanks

Answer 16

Thanks guys!