A094applications


 * Applications of Logarithms in Calculus **


 * Example 1 **

math . \qquad\textbf{a) } \;\; \text{If } \; y = \log_e \big( 3x^2 + 7 \big), \; \text{ Find } \; \dfrac{dy}{dx} \\ . \\ . \qquad \textbf{b) } \;\; \text{Hence find the exact value of }\; \displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx} \qquad. math

__**Solution:**__

math . \qquad\textbf{a) } \;\; y = \log_e \big( 3x^2 + 7 \big) \qquad . \\ . \\ . \qquad \quad \textit{Use the chain rule version of the derivative of a log} \qquad . \\ . \\ . \qquad \quad g(x) = 3x^2 + 7 \; \text{ so } \; g'(x) = 6x \\ . \\ . \qquad \quad \dfrac{dy}{dx} = \dfrac{6x}{3x^2 + 7} \quad \text{ for } \; 3x^2 + 7 > 0 \\ . \\ . \qquad \quad \dfrac{dy}{dx} = \dfrac{6x}{3x^2 + 7} \quad \text { for } \; x \in R math

math . \qquad \textbf{b) } \;\; \textit{Question has } \underline{hence} \textit{ so we } \underline{must} \textit{ use the result from part a} \qquad . math

math . \qquad \text{From part a, we have that:} \\. \\ . \qquad \displaystyle{ \int \dfrac{6x}{3x^2 + 7} \; dx} = \log_e \big( 3x^2 + 7 \big) + c \qquad. math

math . \qquad \text{So} \\. \\ . \qquad \displaystyle{ \int \limits_0^2 \dfrac{6x}{3x^2 + 7} \; dx} = \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2 \qquad. math

math . \qquad \textit{Manipulate this to get the integral we need for the question} \qquad. math

math . \qquad 6 \displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx} = \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2 \qquad. math

math . \qquad \;\; \displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx} = \dfrac{1}{6} \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2 \qquad. math

math . \qquad \qquad \qquad \qquad \quad = \dfrac{1}{6} \Big( \log_e \big( 19 \big) - \log_e \big( 7 \big) \Big) \\. \\ . \qquad \qquad \qquad \qquad \quad = \dfrac{1}{6} \log_e \Big( \dfrac{19}{7} \Big) \quad \text{ square units} \qquad. math


 * Example 2 **

math . \qquad \text {For the function } \; f: R^+ \rightarrow R, \; f(x) = \log_e \big( 2x \big) \\. \\ . \qquad \textbf{a) } \;\; \text{find } \; f^{-1}(x) \; \text{ and sketch the graph of } \; f(x) \; \text{ and } \; f^{-1}(x) \qquad . math

math . \qquad \textbf{b) } \;\; \text{find the exact value of the area given by } \; \displaystyle{ \int \limits_0^{\log_e(4)} f^{-1}(x) \; dx} \qquad . \\ . \\ . \qquad \textbf{c) } \;\; \text{hence find the exact value of } \; \displaystyle{ \int \limits_{\dfrac{1}{2}}^2 f(x)\; dx} math

__**Solution:**__

math . \qquad \textbf{a) } \;\; \text{Given } \; f(x) = \log_e \big( 2x \big) \; \text{ find } \; f^{-1}(x) \qquad . math

math . \qquad \quad \text{Let } \; y = \log_e \big( 2x \big) \\. \\ . \qquad \quad \textit{To find the inverse, swap the x and the y then solve for y} \qquad. \\ . \\ . \qquad \quad x= \log_e \big( 2y \big) \\. \\ . \qquad \quad 2y = e^x \\. \\ . \qquad \quad \; y = \dfrac{1}{2} e^x math

math . \qquad \quad \text{Hence, } \; f^{-1}(x) = \dfrac{1}{2} e^x \quad x \in R \qquad. math

math . \qquad \textbf{a) } \;\; \text{ ... and sketch the graph of } \; f(x) \; \text{ and } \; f^{-1}(x) \qquad . math



math . \qquad \textbf{b) } \;\; \text{find the exact value of the area given by } \; \displaystyle{ \int \limits_0^{\log_e(4)} f^{-1}(x) \; dx} \qquad . math



math . \qquad \quad \displaystyle{ \int \limits_0^{\log_e(4)} f^{-1}(x) \; dx} \qquad. \\ . \\ . \qquad \quad = \displaystyle{ \int \limits_0^{\log_e(4)} \dfrac{1}{2} e^x \; dx} \qquad. \\ . \\ . \qquad \quad = \Big[ \dfrac{1}{2} e^x \Big]_0^{\log_e(4)} math

math . \qquad \quad = \Big( \dfrac{1}{2} e^{log_e(4) } \Big) - \Big( \dfrac{1}{2} e^0 \Big) \qquad. \\ . \\ . \qquad \quad = \Big( \dfrac{1}{2} (4) \Big) - \Big( \dfrac{1}{2} \Big) \\. \\ . \qquad \quad = \dfrac{3}{2} \quad \text{ square units} math

math . \qquad \textbf{c) } \;\; \text{hence find the exact value of } \; \displaystyle{ \int \limits_{\dfrac{1}{2}}^2 f(x)\; dx} \qquad . math

math . \qquad \quad \displaystyle{ \int \limits_{\dfrac{1}{2}}^2 f(x)\; dx} = \displaystyle{ \int \limits_{\dfrac{1}{2}}^2 \log_e \big( 2x \big) \; dx} \qquad. math



... ... ... //We have no way of anti-differentiating a log, so we need to use the result from part (b)}//

... ... ... //In the graph above the red section is the desired area. But on that graph, where would our area for part (b) appear?//



... ... ... //As can be seen, the blue area from part (b) plus the red area from part (c) form a rectangle with a base of 2 and a height of log// e //(4)//

math . \qquad \textbf{Area(} red \textbf{) = Area( } rectangle \textbf{ ) - Area( } blue \textbf{)} \qquad. \\ . \\ . \qquad \qquad \qquad = 2 \times \log_e (4) - \dfrac{3}{2} math

math . \qquad \text{hence} \\. \\ . \qquad \displaystyle{ \int \limits_{\dfrac{1}{2}}^2 \log_e \big( 2x \big) \; dx} = 2\log_e (4) - \dfrac{3}{2} \quad \text{square units} \qquad. math

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