The problem: After 10 days the amount of radioactive substance had reduced to half of the initial amount. In what time will the amount diminish to 5% of the original amount?
I found almost the right answer to the problem only the decimal places don't fit.
My solution: $\frac{N(0)}{N(10)}=e^{-10k}=\frac{1}{2}$ and $\frac{N(0)}{t}=e^{-kt}=\frac{1}{20}.$ From here it follows that $10e^{-kt}=e^{-10k}.$ Taking a natural logarithm of both sides: $-kt \cdot ln(10)= -10k \Rightarrow t= \frac{10}{ln(10)}=10:2,3 = 4,347.$ But the answer given is 43,22. So not only decimal place but the answer itself is incorrect.
You applied the logarithm arithmetic law incorrectly, you should get with your approach $$ -kt+\ln(10)=-10k, $$ where you can not cancel $k$. Instead take the logarithm in both identities separately $$ -10k=-\ln(2),~~ -kt=-\ln(20)\implies t=10\frac{\ln(20)}{\ln(2)}=43.2192809488736. $$