simplify and show without logs problem - again, driving me nuts!!

[han_bozo]

Struggling with the one below. I'm fine with the first few steps but unsure about the last few steps and what transformations to make as well as dealing with that negative sign! Please help!

lnI = ln(2V) - {ln(KR + r) - lnK + KL}


I basically get to:

ln(2V)
divided by
ln(K(KR + r) - KL

If I take logs, I'm not sure how to deal with the -lnKL??? I think it could be related to it being transformed to: ln(base KL)e so it'll go in the numerator right? but what about the neg sign? :S

 

[Deleted member 4993]

Struggling with the one below. I'm fine with the first few steps but unsure about the last few steps and what transformations to make as well as dealing with that negative sign! Please help!

lnI = ln(2V) - {ln(KR + r) - lnK + KL}


I basically get to:

ln(2V)
divided by
ln(K(KR + r) - KL

If I take logs, I'm not sure how to deal with the -lnKL??? I think it could be related to it being transformed to: ln(base KL)e so it'll go in the numerator right? but what about the neg sign? :S

Assuming you written the problem correctly:

Put all the logs on one side (and light a match to it)):

lnI = ln(2V) - {ln(KR + r) - lnK + KL}

lnI = ln(2V) - ln(KR + r) + lnK - KL

lnI - ln(2V) + ln(KR + r) - lnK = -KL

\(\displaystyle ln\dfrac{I*(KR+r)}{2KV} \ = \ -KL \)

\(\displaystyle \dfrac{I*(KR+r)}{2KV} \ = \ e^{-KL} \)

 

[stapel]

lnI = ln(2V) - {ln(KR + r) - lnK + KL}

Would it be correct to assume that you are supposed to solve the above equation for one of the variables?

Also, is the following a correct statement of the equation?

. . . . .\(\displaystyle \ln(I)\, =\, \ln(2V)\, -\, \bigg(\, \ln(KR\, +\, r)\, -\, \ln(K)\, +\, KL\, \bigg)\)

In other words, is it correct that the "KL" is not included inside the argument of a logarithm?

Thank you!

 

[han_bozo]

Many thanks Subhotosh Khan! Out of interest, how does '-lnKL' turn into '-e to the power of 'KL'?? <-- this is probably the only thing I'm struggling with!

With your solution though, I think it's a little off to what I was expecting.

Hi Stapel! Yes, what you've written is correct. Once logs have been taken, the only log left to deal with is lnKL located in the denominator. I'm expecting an answer of something like the one below but I'm not sure how the numerator is obtained!

I = 2Ve^KL/ K(KR+r)


I think it has to do with some manipulation of 1/lnKL but this is negative and it could be re-written as 'ln_KLe'??? and then some sort of further manipulation needed? *sighs*

Many thanks all!!!

 

[Deleted member 4993]

Many thanks Subhotosh Khan! Out of interest, how does '-lnKL' turn into '-e to the power of 'KL'?? <-- this is probably the only thing I'm struggling with!

With your solution though, I think it's a little off to what I was expecting.

Hi Stapel! Yes, what you've written is correct. Once logs have been taken, the only log left to deal with is lnKL located in the denominator. I'm expecting an answer of something like the one below but I'm not sure how the numerator is obtained!

I = 2Ve^KL/ K(KR+r)


I think it has to do with some manipulation of 1/lnKL but this is negative and it could be re-written as 'ln_KLe'??? and then some sort of further manipulation needed? *sighs*

Many thanks all!!!

If you want that answer then your starting equation should be (instead of lnI = ln(2V) - {ln(KR + r) - lnK + KL}):

lnI = ln(2V) - {ln(KR + r) + lnK} + KL

 

[han_bozo]

That was how the question was presented to me so I've only reproduced it exactly how it was. Thanks.