taking the natural log of $\mathrm{e}^{2x} =\frac{4}{3} $ I have been unable to answer the following question.
I must solve for $x$:
$$e^{2x} = \frac{4}{3}$$
I have been made aware that I must take the natural logarithm at both sides, giving:
$$\ln(\mathrm{e}^{2x}) = \ln\left( \frac{4}{3} \right)$$
The solution is : $$x = \frac{1}{2} \cdot \ln\left( \frac{4}{3} \right)$$
I am not sure how my tutor has arrived at this point.
If anyone could show me how to solve it giving more steps then it would be very much appreciated. thanks.
 A: Note that
$\ln(a^x)=x\ln(a)$
Thus
$$ln(e^{2x}) = ln(4/3)$$
$$(2x)ln(e) = ln(4/3)$$
$$2x = ln(4/3)$$
since $\ln(e)=1$
$$x=\frac{1}{2}\ln(\frac{4}{3})$$
A: Notice, a few things:


*

*$$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$

*$$\ln(e)=\log_e(e)=\frac{\ln(e)}{\ln(e)}=1$$

*$$\ln(x)=\log_e(x)=\frac{\ln(x)}{\ln(e)}=\frac{\ln(x)}{1}=\ln(x)$$

*$$\ln(a^x)=x\ln(a)\space\space\space\text{when}\space a,x\space\text{are positive}$$

*$$\ln\left(\frac{a}{x}\right)=\ln(a)-\ln(x)\space\space\space\text{when}\space a,x\space\text{are positive}$$


So, solving your question:
$$e^{2x}=\frac{4}{3}\Longleftrightarrow$$
$$\ln\left(e^{2x}\right)=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$
$$2x\ln\left(e\right)=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$
$$2x\cdot1=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$
$$2x=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$
$$x=\frac{\ln\left(\frac{4}{3}\right)}{2}\Longleftrightarrow$$
$$x=\frac{\ln(4)-\ln(3)}{2}\Longleftrightarrow$$
$$x=\frac{\ln(2^2)-\ln(3)}{2}\Longleftrightarrow$$
$$x=\frac{2\ln(2)-\ln(3)}{2}\Longleftrightarrow$$
$$x=\ln(2)-\frac{\ln(3)}{2}\approx0.1438410362$$
