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Question

Question: How do you evaluate \[{\log _\pi }e?\]...

How do you evaluate logπe?{\log _\pi }e?

Explanation

Solution

In this type of question we will use logarithmic and exponential base change properties to solve the question. According to base change formula the base of a logarithmic function can be changed as logax=logcxlogca{\log _a}x = \dfrac{{{{\log }_c}x}}{{{{\log }_c}a}} , where aa and xx are all positive real numbers. Use the above base change formula and we get the required answer.

Complete step by step answer:
Given logπe{\log _\pi }e
We use the formula of logarithmic function we get
logπe=logelogπ{\log _\pi }e = \dfrac{{{{\log }}e}}{{{{\log }}\pi }}
Putting the values of numerator and denominator , we get
logπe=0.43429448190.49714987269\Rightarrow {\log _\pi }e = \dfrac{{0.4342944819}}{{0.49714987269}}
Calculating and simplifying in the above equation and we get
logπe=0.873568526\Rightarrow {\log _\pi }e = 0.873568526
We take approximation of the above value and we get
logπe=0.87357\Rightarrow {\log _\pi }e = 0.87357 (approximately)

Note: We should use the value of logeπ=0.49714987269{\log _e}\pi = 0.49714987269 and logee=0.4342944819{\log _e}e = 0.4342944819 in the function to find the value of yy. We know the power rule of logarithm log(ab)=b.loga\log ({a^b}) = b.\log a, the quotient rule of logarithm log(ab)=logalogb\log \left( {\dfrac{a}{b}} \right) = \log a - \log b and the product rule of logarithm log(a.b)=loga+logb\log \left( {a.b} \right) = \log a + \log b. We use these depending on the problem and we should know that while applying these laws the base of the logarithm should be the same.