# Use the properties of logarithms to write the expression as a single term

I am asked to use the properties of logarithms to write the following expression as a single term:

$$(1/2)\ln(4t^4) - \ln b$$

I have the solution here but I get stumped halfway through:

$$(1/2)\ln(4t^4) - \ln b$$

$$= \ln (4t^4)^{(1/2)} - \ln b$$

$$\mathbf{4t^4 = (2t^2)^2 => }$$

$$\mathbf{ = \ln ((2t^2)^2)^{(1/2)} - \ln b }$$

I don't quite understand how that transformation (the lines I bolded) takes place? I get that the first $$4$$ can turn into $$2^2$$, but what happened to the exponent of $$4$$? Why did that get turned into a $$2$$?

$$= \ln (2t^2) - \ln b$$

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

Note that $$4t^4=4(t^4),$$ not $$(4t)^4$$. Now, using the properties of exponents $$(2t^2)^2=2^2(t^2)^2=4t^4$$
$$\begin{array}\\ \frac12\ln(4t^4) - \ln b &=\ln((4t^4)^{1/2}) - \ln b \qquad u\ln(v) = \ln(v^u)\\ &=\ln(2t^2) - \ln b\\ &=\ln\left(\dfrac{2t^2}{b}\right) \qquad \ln(u)-\ln(v) = \ln(\frac{u}{v})\\ \end{array}$$