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.
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:
$$ \ln \frac{12}{1 +e^{-0.6(t-6)}} = \ln 12 - \ln(1 + e^{-0.6(t-6)})$$
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.
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 $$t=6-\frac 53 \log\left(\frac{12}{B(t)}-1 \right)$$
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.
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.
$\frac65=1+x$
$x=\frac15$
$e^{-0.6(t-6)}=\frac15$
$-0.6(t-6)=ln(0.2)$
$0.6(t-6)=-ln(0.2)$
$(t-6)=\frac{-ln(0.2)}{0.6}$
$t=\frac{-ln(0.2)}{0.6}+6=8.6824$
