Question 235064
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a.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\ln(2)\ +\ \ln(x^2)\ +\ 2]


The first thing to do is look at the definition of a logarithm.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ b^y = x \Rightarrow\ \ y = \log_b(x)]  


The base of the *[tex \Large \ln] function is *[tex \Large e], so we can say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ e^2 = x \Rightarrow\ \ 2 = \log_e(e^2)]


So we can now re-write your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\ln(2)\ +\ \ln(x^2)\ +\ \ln(e^2)]


Now use:


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


To say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(2^3)\ +\ \ln(x^2)\ +\ \ln(e^2)]


and simplify to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(8)\ +\ \ln(x^2)\ +\ \ln(e^2)]


Now use the property that the sum of the logs is the log of the product, that is:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\ln(2)\ +\ \ln(x^2)\ +\ 2\ =\ \ln(8e^2x^2)\ \approx \ln(59.11x^2)]


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b.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\,\cdot\,\log_a(4)\ -\ 2\,\cdot\,\log_a(5)]


Use


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


To say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_a(4^{\frac{1}{2}})\ -\ \log_a(5^2)]


Recall that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{\frac{1}{n}} = \sqrt[n]{a}]


To say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_a(2)\ -\ \log_a(25)]


Now use the property that 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)]


To say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\,\cdot\,\log_a(4)\ -\ 2\,\cdot\,\log_a(5)\ =\ \log_a\left(\frac{2}{25}\right)]



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