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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} $

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2 Answers 2

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Note that $4t^4=4(t^4),$ not $(4t)^4$. Now, using the properties of exponents $$(2t^2)^2=2^2(t^2)^2=4t^4$$

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$\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} $

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