Solving $0.012 = 0.5e^x$ gives me two different answers

Question

I was performing a calculation when I reached this point of the calculation.

$$0.012 = 0.5e^x$$

I thought I could do the following

$$\frac{0.012}{0.5} = e^x= 0.024$$

hence $x = \ln(0.024) = -3.73$ to 3 s.f.

The book, however required a positive integer however.

Its working is shown as below.

$$0.012 = 0.5e^x$$ $$\ln(0.012) = \ln(0.5)x$$ $$\frac{\ln(0.012)}{\ln(0.5)} = x = 6.380... $$

hence giving an answer of $6.38$ to $3$ s.f.

I am struggling to see what I have done wrong.

Can someone help?

Thanks.

Answer 1

The solution given in your book is incorrect $$0.012 = 0.5e^x$$ $$\implies \ln(0.012) = \ln(0.5) \color{blue}+ x \color{red}\neq \ln(0.5)x$$

Note that $e^x$ is one-one function I'd est, it can't have same value for different $x$

(Where $x \in \mathbb R$)

Answer 2

Notice $\ln(xy) = \ln(x) + \ln(y)$

$$0.012 = 0.5e^x$$ $$\ln(0.012) =\ln(0.5) + \ln(e^x) = \ln(0.5)+ x$$