Question 606679
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I'm not sure whether you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\sqrt{\frac{a}{c}}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\frac{\sqrt{a}}{c}\right)]


Either way, proceed as follows:


Recall that the log of the quotient is the difference of the logs, so you either have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\sqrt{a}\right)\ -\ \log_b\left(\sqrt{c}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\sqrt{a}\right)\ -\ \log_b\left(c\right)]


Then recall that *[tex \LARGE \sqrt{\alpha}\ =\ \alpha^{\frac{1}{2}}]  and that *[tex \LARGE \log_b\left(\alpha^n\right)\ =\ n\,\cdot\,\log\left(\alpha\right)]


So you either have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\,\cdot\,\log_b\left(a\right)\ -\ \frac{1}{2}\,\cdot\,\log_b\left(c\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\,\cdot\,\log_b\left(a\right)\ -\ \log_b\left(c\right)]


Substituting, you either have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ 2\ =\ 3]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ 4\ =\ 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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