# How do you simplify 1/2 ln (4t^4) - ln 2?

May 4, 2016

This is asking you to remember the properties of logarithms. Here are the ones you need to know:

• $\setminus m a t h b f \left(\ln {a}^{b} = b \ln a\right)$
• $\setminus m a t h b f \left(c \ln a - c \ln b = c \ln \setminus \frac{a}{b}\right)$

So, we can start by getting that exponent out in front:

$\frac{1}{2} \ln 4 {t}^{4} - \ln 2$

$= \frac{1}{2} \ln {\left(2 {t}^{2}\right)}^{2} - \ln 2$

Be careful that you do the above step correctly. It would be incorrect to change $\ln 4 {t}^{4}$ to $4 \ln 4 t$, because the exponent only applied to the $t$ at that time.

(If you did, you would imply that the expression was $\ln {\left(4 t\right)}^{4} = \ln 256 {t}^{4} \ne \ln 4 {t}^{4}$.)

Now, the exponent applies to the quantity $2 {t}^{2}$, so we are justified in moving the power out to the front as a coefficient!

$= \cancel{\frac{1}{2}} \cdot \cancel{2} \ln 2 {t}^{2} - \ln 2$

$= \ln 2 {t}^{2} - \ln 2$

With the same coefficients $c = 1$ in front, we can now turn this into a fraction:

$= \ln \setminus \frac{\cancel{2} {t}^{2}}{\cancel{2}}$

$= \ln {t}^{2}$

$= \textcolor{b l u e}{2 \ln t}$