09 3applications

# Applications of Logarithms in Calculus

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### Example 1

a) .. If $y = \log_e \big( 3x^2 + 7 \big)$, Find $\dfrac{dy}{dx}$

b) .. Hence find the exact value of $\displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx}$

Solution:
a) .. If $y = \log_e \big( 3x^2 + 7 \big)$, Find $\dfrac{dy}{dx}$
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… … … Use the chain rule version of the derivative of a log
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… … … $g(x) = 3x^2 + 7 \quad \text{ so } \; g'(x) = 6x$
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… … … $\dfrac{dy}{dx} = \dfrac{6x}{3x^2 + 7} \quad \text{ for } \; 3x^2 + 7 > 0$
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… … … $\dfrac{dy}{dx} = \dfrac{6x}{3x^2 + 7} \quad \text { for } \; x \in R$

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b) .. Hence find the exact value of $\displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx}$

… … … Question has hence so we must use the result from part a
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… … … From part a, we have that:
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… … … $\dfrac{d}{dx} \Bigg(\log_e \big( 3x^2 + 7 \big) \Bigg) = \dfrac{6x}{3x^2 + 7}$
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… … … So
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… … … $\displaystyle{ \int \dfrac{6x}{3x^2 + 7} \; dx} = \log_e \big( 3x^2 + 7 \big) + c$
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… … … So
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… … … $\displaystyle{ \int \limits_0^2 \dfrac{6x}{3x^2 + 7} \; dx} = \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2$
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… … … Manipulate this to get the integral we need for the question
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… … … $6 \displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx} = \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2$
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… … … $\;\; \displaystyle{ \int \limits_0^2 \dfrac{x}{3x^2 + 7} \; dx} = \dfrac{1}{6} \Big[ \log_e \big( 3x^2 + 7 \big) \Big]_0^2$
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… … … … $= \dfrac{1}{6} \Big( \log_e \big( 19 \big) - \log_e \big( 7 \big) \Big)$
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… … … … $= \dfrac{1}{6} \log_e \Big( \dfrac{19}{7} \Big)$ square units

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### Example 2

For the function $\; f: R^+ \rightarrow R, \; f(x) = \log_e \big( 2x \big)$

a) .. find $f^{-1}(x)$ and sketch the graph of $f(x) \; \text{ and } \; f^{-1}(x)$

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

c) .. hence find the exact value of $\displaystyle{ \int \limits_{\frac{1}{2}}^2 f(x)\; dx}$

Solution:

a) .. find $f^{-1}(x)$ and sketch the graph of $f(x) \; \text{ and } \; f^{-1}(x)$
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… … … Let $y = \log_e \big( 2x \big)$
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… … … To find the inverse, swap the x and the y then solve for y
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… … … $x= \log_e \big( 2y \big)$
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… … … $2y = e^x$
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… … … $\; y = \dfrac{1}{2} e^x$
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… … … Hence, $\; f^{-1}(x) = \dfrac{e^x}{2} \quad x \in R$
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… … a) .. … and sketch the graph of $f(x) \; \text{ and } \; f^{-1}(x)$

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

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… … … $\displaystyle{ \int \limits_0^{\log_e(4)} f^{-1}(x) \; dx}$
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… … … $= \displaystyle{ \int \limits_0^{\log_e(4)} \dfrac{1}{2} e^x \; dx}$
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… … … $= \Big[ \dfrac{1}{2} e^x \Big]_0^{\log_e(4)}$
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… … … $= \Big( \dfrac{1}{2} e^{log_e(4) } \Big) - \Big( \dfrac{1}{2} e^0 \Big)$
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… … … $= \Big( \dfrac{1}{2} (4) \Big) - \Big( \dfrac{1}{2} \Big)$
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… … … $= \dfrac{3}{2}$ square units

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c) .. hence find the exact value of $\displaystyle{ \int \limits_{\frac{1}{2}}^2 f(x)\; dx}$
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… … … $\displaystyle{ \int \limits_{\frac{1}{2}}^2 f(x)\; dx} = \displaystyle{ \int \limits_{\frac{1}{2}}^2 \log_e \big( 2x \big) \; dx}$
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… … … We have no way of anti-differentiating a log, so we need to use the result from part b

… … … The question included hence, so that is a clue.
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… … … In the graph above, the red section is the desired area.

… … … But on that graph, where would our area for part b appear?
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… … … 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 $\log_e(4)$
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… … … Area( red ) = Area( rectangle ) $-$ Area( blue )
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… … … Area( red ) $= 2 \times \log_e (4) - \dfrac{3}{2}$
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… … … Area( red ) $= 2 \times \log_e (2^2) - \dfrac{3}{2}$
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… … … Area( red ) $= 4 \times \log_e (2) - \dfrac{3}{2}$
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… … … hence
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… … … $\displaystyle{ \int \limits_{\frac{1}{2}}^2 \log_e \big( 2x \big) \; dx} = 4\log_e (2) - \dfrac{3}{2}$ square units

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