Question 328991
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 144\ =\ 16\ \cdot\ 9\ =\ 2^4\ \cdot\ 3^2]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(144)\ =\ 4\log_x(2)\ +\ 2\log_x(3)\ =\ 0.56\ +\ 0.256\ =\ 0.816]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(x^2)\ =\ 2\log_x(x)]


but *[tex \LARGE \forall b\ \in\ \mathbb{R}, b\ >\ 0, \log_b(b)\ =\ 1].


Since *[tex \LARGE \log_x(2)\ \in\ \mathbb{R}], we are certain that *[tex \LARGE x\ \in\ \mathbb{R}, x\ >\ 0], therefore *[tex \LARGE \log_x(x)\ =\ 1]


So we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(x^2)\ =\ 2\log_x(x)\ =\ 2]


And finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25\ =\ 5^2]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(25)\ =\ 2\log_x(5)\ =\ 0.008]


The log of the product is the sum of the logs and the log of the quotient is the difference of the logs, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x\left(\frac{144x^2}{25}\right)\ =\ .818\ +\ 2\ -\ 0.008\ =\ 2.808]



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