Logarithmic equation problem

Question

I’m working through an old copy of K.A.Stroud, Engineering Mathematics (5th ed) and one of the exercises has the following logarithm problem where the aim is to re-write without logs:

$\ln{I} = \ln{(2V)} - \{\ln{(KR + r)} - \ln{K} + KL\}$

The provided answer is:

$I = \frac{2Ve^{KL}}{K(KR+r)}$

I can’t quite get there as I end up with the e term in the denominator. Would be great to see a worked solution if anyone has the time.

Thanks in advance.

Answer 1

You're right about the $e$ term in the denominator.

Simplifying,

$$\ln{I}=\ln{2V}-\{\ln(KR+r)-\ln{K}+KL\}$$ $$\implies \ln{I}=\ln{2V}-\ln(KR+r)+\ln{K}-KL$$ $$\implies \ln{I}=\ln{2V}-\ln(KR+r)+\ln{K}-\ln{e^{KL}}$$ $$\implies \ln{I}=\ln\left({2V}{\frac{1}{KR+r}}{K}{\frac{1}{e^{KL}}}\right)$$ $$\implies \ln{I}=\ln\left(\frac{2VK}{(KR+r)e^{KL}}\right)$$ $$\implies I=\frac{2VK}{(KR+r)e^{KL}}$$

Answer 2

I think you may have made a typo; I hope the original expression was $$\ln I = \ln(2V) - (\ln(KR+r)+\ln(K)-KL)$$

Then we get $$\begin{aligned}I &= e^{\ln(2V)-(\ln(KR+r)+\ln(K)-KL)}\\ &=e^{\ln(2V)}e^{-(\ln(KR+r)+\ln(K)-KL)}\\&=\dfrac{e^{\ln(2V)}}{e^{(\ln(KR+r)+\ln(K)-KL)}}\\&=\dfrac{2V}{e^{(\ln(KR+r)+\ln(K)-KL)}}\\&=\dfrac{2V}{e^{(\ln(KR+r)}e^{\ln(K)}e^{-KL}}\\&=\dfrac{2V}{(KR+r)(K)\dfrac{1}{e^{KL}}}\\&=\dfrac{2Ve^{KL}}{K(KR+r)} \end{aligned}$$