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2.4: General Exponential and Logarithmic Functions

  • Page ID
    54772
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    For a general exponential function \(y = a^x\), with \(a > 0\), use logarithmic differentiation to find its derivative:

    \[\begin{aligned} \ln\,y ~&=~ \ln\,\left(a^x\right) ~=~ x \,\ln\,a\\ \ddx\,(\ln\,y) ~&=~ \ddx\,(x \,\ln\,a) ~=~ \ln\,a\\ \frac{y'}{y} ~&=~ \ln\,a \quad\Rightarrow\quad y' ~=~ y \cdot \ln\,a\end{aligned} \nonumber \]

    Thus, the derivative of \(y = a^x\) is: In general, for an exponent of the form \(u = u(x)\):

    Find the derivative of \(y = 2^{\cos x}\).

    Solution: This is the case where \(a = 2\), so:

    \[\dydx = (\ln\,2)\;2^{\cos x} \;\cdot\; \ddx\,(\cos x) ~=~ -(\ln\,2)\;(\sin x)\;2^{\cos x} \nonumber \]

    Note that any exponential function \(y = a^x\) can be expressed in terms of the exponential function \(e^x\). Since

    \[a^x ~>~ 0 \quad\Rightarrow\quad e^{\ln\,(a^x)} ~=~ a^x ~~, \nonumber \]

    and since \(\ln\,(a^x) = x\,\ln\,a\), then: Computers and calculators often use the above formula to calculate \(a^x\).

    The function \(y = a^x\) has an inverse for any \(a > 0\), except for \(a = 1\) (in that case \(y = 1^x = 1\) is just a constant function). To see this, notice that since \(a^x > 0\) for all \(x\), and \(\ln\,a < 0\) for \(0 < a < 1\), while \(\ln\,a > 0\) for \(a > 1\), then \(\dydx = (\ln\,a)\,a^x\) is always negative if \(0 < a < 1\) and always positive if \(a > 1\). Thus, \(y = a^x\) is a strictly decreasing function if \(0 < a < 1\), and it is a strictly increasing function if \(a > 1\). The graphs in each case are shown in Figure [fig:expa].

    Hence, for any \(a >0\) with \(a \ne 1\) the function \(f(x) = a^x\) is one-to-one, so it has an inverse function, called the base \(\bm{a}\) logarithm and denoted by \(f^{-1}(x) = \log_a x\). It is often spoken as “log base \(a\) of \(x\)”. The graphs for \(a < 1\) and \(a > 1\) are shown in Figure [fig:loga]. Note that the natural logarithm is just the base \(a\) logarithm in the special case with \(a = e\), i.e. \(\ln\,x = \log_e x\). The base \(a\) logarithm has properties similar to those of the natural logarithm (and the corresponding properties of \(a^x\)):

    Note that \(\log_a x\) can be put in terms of the natural logarithm, since

    \[x ~=~ a^{\log_a x} \quad\Rightarrow\quad \ln\,x ~=~ \ln\,\left(a^{\log_a x}\right) ~=~ (\log_a x) \cdot (\ln\,a) \nonumber \]

    so dividing the last expression by \(\ln\,a\) gives: The above formula is useful on calculators that do not have a \(\log_a x\) key or function. Taking the derivative of both sides yields: In general, when taking the logarithm of a function \(u = u(x)\):

    Find the derivative of \(y = \log_2 (\cos\,4x)\).

    Solution: This is the case where \(a = 2\), so:

    \[\dydx = \frac{1}{(\cos\,4x)\,(\ln\,2)} \;\cdot\; \ddx\,(\cos\,4x) ~=~ -\frac{4 \sin\,4x}{(\ln\,2)\,(\cos\,4x)} \nonumber \]

    The number \(a\) is the base of both the logarithm function \(\log_a x\) and the exponential function \(a^x\). Base 2 and base 10 are the most commonly used bases other than base \(e\). Base 10 is how numbers are normally expressed, as combinations of powers of 10 (e.g. \(2014 = \bm{2} \cdot 10^3 \;+\; \bm{0} \cdot 10^2 \;+\; \bm{1} \cdot 10^1 \;+\; \bm{4} \cdot 10^0\)). Base 2 is especially useful in computer science, since computers represent all numbers in binary format, i.e. as a sequence of zeros and ones, indicating how many successive powers of two to take and then sum up.8 For example, the number 6 is represented in binary format as 110, since \(\bm{1} \cdot 2^2 \;+\; \bm{1} \cdot 2^1 \;+\; \bm{0} \cdot 2^0 ~=~ 4 + 2 + 0 = 6\).

    [sec2dot4]

    For Exercises 1-9, find the derivative of the given function.

    3

    \(y = \dfrac{3^x ~+~ 3^{-x}}{2}\)

    \(y = 2^{\ln\,3x}\vphantom{\dfrac{3^x}{2}}\)

    \(y = 2^{2^x}\vphantom{\dfrac{3^x}{2}}\)

    3

    \(y = \tan^{-1} \pi^x\)

    \(y = \log_2 \,(x^2 + 1)\)

    \(y = \log_{10} \,e^x\)

    3

    \(y = \sin\,\left(\log_2 \,\pi x\right)\)

    \(y = \log_2 \,4^{2x}\)

    \(y = 8^{\log_2 \,x}\)

    [[1.]]

    Show that for all constants \(k\) the function \(y = A a^{\frac{kx}{\ln\,a}}\) satisfies the differential equation \(\dydx = ky\). Does this contradict the statement made in Section 2.3 that the only solution to that differential equation is of the form \(y = A e^{kx}\)? Explain your answer.


    This page titled 2.4: General Exponential and Logarithmic Functions is shared under a GNU General Public License 3.0 license and was authored, remixed, and/or curated by Michael Corral.

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