taking the natural log of $\mathrm{e}^{2x} =\frac{4}{3} $

Question

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.

Answer 1

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:

$$e^{2x}=\frac{4}{3}\Longleftrightarrow$$ $$\ln\left(e^{2x}\right)=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$ $$2x\ln\left(e\right)=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$ $$2x\cdot1=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$ $$2x=\ln\left(\frac{4}{3}\right)\Longleftrightarrow$$ $$x=\frac{\ln\left(\frac{4}{3}\right)}{2}\Longleftrightarrow$$ $$x=\frac{\ln(4)-\ln(3)}{2}\Longleftrightarrow$$ $$x=\frac{\ln(2^2)-\ln(3)}{2}\Longleftrightarrow$$ $$x=\frac{2\ln(2)-\ln(3)}{2}\Longleftrightarrow$$ $$x=\ln(2)-\frac{\ln(3)}{2}\approx0.1438410362$$

Answer 2

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 $\ln(e)=1$ $$x=\frac{1}{2}\ln(\frac{4}{3})$$