Law of Natural Logarithms

Question

This is an old example and since I've free time, I am working on it.

$$B(t)= \frac{12}{1+e^{-0.6(t-6)}} $$

If we set $$10= \frac{12}{1+e^{-0.6(t-6)}}$$ $$ \ln 10 = \ln \frac{12}{1+e^{-0.6(t-6)}} $$ We know that:$ \ln 1 = 0 $ $$ \ ln (e^r)= r $$

So $$ \ln 10 = \frac{\ln 12}{( -0.6(t -6) )}$$

$$t=\frac{ \ln (\frac{10}{12-10})}{ (0.6) } +6 $$ $$t= 8.682 $$

What's wrong?

A good example of this problem might be:

$P = \frac{k}{ 1+ e^{-rt} } $ for $ t$

Thank you in advance.

Answer 1

The following two steps are not legal operations involving the logarithm function.

1) From $\ln\frac{12}{1+e^r}$ to $\frac{\ln12}{\ln{1+e^r}}$

2) From $\ln{1+e^r}$ to $\ln1+\ln e^r$

You can only use the logarithm rules you have learnt in class:

A) $\ln (a\times b)=\ln a+\ln b$

B) $\ln \frac{a}{b} = \ln a - \ln b$

C) $\ln a^n = n\ln a$

D) $\log_x a=\frac{\log_y a}{\log_y x}$ (You may not have learnt this one yet.)

The correct way to solve your original problem is as follows:

$$10=\frac{12}{1+e^{-0.6(t-6)}}$$

Divide by 10 and multiple by $1+e^{-0.6(t-6)}$

$$1+e^{-0.6(t-6)}=\frac{6}{5}$$

Subtract $1$

$$e^{-0.6(t-6)}=\frac{1}{5}$$

Now convert to a logarithm

$$-0.6(t-6)=\ln\frac{1}{5}$$

Rearrange to get $t$

$$t-6 = -\frac{5}{3}\ln5^{-1}$$

$$t=6+\frac{5}{3}\ln5$$

Note in the last step the negative power was moved to the front using the rule I listed as C.

Answer 2

Logarithms do not "distribute" over multiplication and division like that. The two rules are this:

$$\ln(AB) = \ln A + \ln B$$

and

$$\ln\left(\frac AB\right) = \ln A - \ln B$$

It appears that you tried to do this:

$$ \ln \frac{12}{1 +e^{-0.6(t-6)}} = \frac{\ln 12}{\ln 1 + \ln e^{-0.6(t-6)}}$$

But that is not correct. The best we can do is this:

$$ \ln \frac{12}{1 +e^{-0.6(t-6)}} = \ln 12 - \ln(1 + e^{-0.6(t-6)})$$

Answer 3

It seems that there are quite many mistakes in your solution.

Let me start from scratch. You have $$B(t)= \frac{12}{1+e^{-0.6(t-6)}}$$ and you want to compute $t$ for a given value of $B(t)$. The steps could be $$\frac{B(t)}{12}= \frac{1}{1+e^{-0.6(t-6)}}\implies \frac{12}{B(t)}={1+e^{-0.6(t-6)}}\implies \frac{12}{B(t)}-1=e^{-0.6(t-6)}$$ Now, take logarithms $$-0.6(t-6)=\log\left(\frac{12}{B(t)}-1 \right)$$ from which $$t=6-\frac 53 \log\left(\frac{12}{B(t)}-1 \right)$$

Answer 4

You have

$$10= \frac{12}{1+e^{-0.6(t-6)}}$$

Then you take ln on both sides.

$$ \ln 10 = \ln \frac{12}{1+e^{-0.6(t-6)}}=ln\frac{A}B $$

But then you transform the RHS from $ln\frac{A}B $ int0 $\frac{\ln A}{\ln B}$. This is an invalid transformation.

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The right way is to solve for $x=e^{-0.6(t-6)}$

$10= \frac{12}{1+x}$

Dividing both sides by $12$

$\frac{10}{12}= \frac{1}{1+x}$

Taking the reciprocal on both sides.

$\frac65=1+x$

$x=\frac15$

$e^{-0.6(t-6)}=\frac15$

$-0.6(t-6)=ln(0.2)$

$0.6(t-6)=-ln(0.2)$

$(t-6)=\frac{-ln(0.2)}{0.6}$

$t=\frac{-ln(0.2)}{0.6}+6=8.6824$