Question 242767
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Sorry, but #1 is confusing.  What part goes inside of the radical?


The way to represent square root with the keyboard is:  sqrt(x) which means *[tex \LARGE \sqrt{x}].  Everything you put inside of the parentheses is under the radical.  The way you wrote your problem, I can't tell if you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_c\left(3\frac{\sqrt{y^3}z^2}{x^4}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_c\left(3\frac{\sqrt{y^3z^2}}{x^4}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_c\left(3\sqrt{\frac{y^3z^2}{x^4}}\right)]


And you don't actually say what it is you want to do with this logarithm.


#2.  Always put your log arguments in parentheses.  I presume you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\log(a)\ \ \log(2)]


I can only presume, because you didn't take the time to say, that you want to express this as a single logarithm.


First:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n) = n\log_b(x)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\log(a)\ -\ \log(2)\ =\ \log\left(a^{\frac{1}{2}}\right)\ -\ \log(2)\ =\ \log\left(\sqrt{a}\right)\ -\ \log(2)]


The difference of the logs is the log of the quotient, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\sqrt{a}\right) - \log(2) = \log\left(\frac{\sqrt{a}}{2}\right)]


Use the same rules for solving 3 as I showed you for 2 with the addition of the rule:


The sum of the logs is the log of the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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