Laws of Exponential and Logarithmic Functions

Laws, Properties, and Derivatives

Laws of Exponential and Logarithmic Functions

Let \(a\) and \(b\) be positive numbers and \(x\) and \(y\) be real numbers. Let \(m\) and \(n\) be positive numbers.

Exponential

• Addition of Exponents Law

  \[b^x b^y = b^{x + y}\]

• Difference of Exponents Law

  \[\frac{b^x}{b^y} = b^{x - y}\]

• Exponentiation Law

  \[\left( b^x \right)^y = b^{xy}\]

• Product Distribution Law

  \[(ab)^x = a^xb^x\]

• Fractional Distribution Law

  \[\left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}\]

Logarithmic

• Logarithmic Addition Law

  \[\log_{b}(mn) = \log_b(m) + \log_b(n)\]

• Logarithmic Subtraction Law

  \[\log_{b}\left( \frac{m}{n} \right) = \log_b(m) - \log_b(n)\]

• Logarithm of a Power

  \[\log_b(m^n) = n \log_b(m)\]

• Logarithm of 1

  \[\log_b(1) = 0 ~~\&~~ \ln(1) = 0\]

• Logarithm of the Base

  \[\log_b(b) = 1 ~~\&~~ \ln(e) = 1\]

Cancellation Properties of \(e^x\) and \(\ln(x)\)

• For all \(x>0\)

  \[e^{\ln(x)} = x\]

• For all \(x\)

  \[\ln(e^x) = x\]

Derivatives of \(e^x\) and \(\ln(x)\)

• Derivative of \(e^x\)

  \[\frac{d}{dx}e^x = e^x\]

• Derivative of \(e^{f(x)}\)

  \[\frac{d}{dx}e^{f(x)} = e^{f(x)} f'(x)\]

• Derivative of \(\ln(x)\)

  \[\frac{d}{dx}\ln(x) = \frac{1}{x}\]

• Derivative of \(\ln(f(x))\)

  \[\frac{d}{dx}\ln(f(x)) = \frac{1}{f(x)} f'(x)\]

Example 1

Solve for unknown value in the exponent

Find all values of \(x\) such that \(\displaystyle 4^{x-x^2} = \frac{1}{16^x}\).

Step 1:   Write both sides of the equation as a power of   \(4\).

Since the left-hand side is already written as a power of \(4\), focus on the right-hand side.

\[\begin{align*} \frac{1}{16^x} &= 16^{-x} \\ &= (4^2)^{-x} && \hbox{since $16 = 4^2$}\\ &= 4^{-2x} && \hbox{since $(b^x)^y = b^{xy}$} \end{align*}\]

Therefore, the original equation can be rewritten in the following manner:

\[4^{x-x^2} = 4^{-2x}.\]

Step 2:   Set the exponents equal to each other, and solve for   \(x\).

\[\begin{align*} x-x^2 &= -2x && \\ 3x-x^2 &= 0 && \text{move all variables to one side}\\ x(3-x) &= 0 && \text{factor}\\ x = 0&, x = 3 && \text{folve for $x$ by setting each factor equal to $0$} \end{align*}\]

Example 2

Solve for unknown value in the exponent

Find all values of \(x\) such that:

\[2^{2x}-40\cdot 2^x +256 = 0.\]

Step 1:   Rewrite the equation in terms of   \(2^x\).

\[(2^x)^2 -40\cdot 2^x+256 = 0.\]

Step 2:   Let   \(u = 2^x\).

\[u^2-40u+256 =0.\]

Step 3:   Factor and solve for   \(u\).

\[(u-8)(u-32) = 0.\]

\[u=8 ~~~~ \hbox{ or } ~~~~ u=32.\]

Step 4:   Substitute   \(2^x\)   back in for   \(u\)   and solve for   \(x\).

\[\begin{align*} 2^x &= 8 & 2^x &= 32\\ x&=3 & x&=5 \end{align*}\]

Example 3

Solve for unknown value in the exponent

Find all values of \(t\) such that \(\dfrac{360}{1+9e^{-2t}} = 90\).

Step 1:   Isolate   \(e^{-2t}\)   using the following steps.

\[\begin{align*} 360 &= 90(1+9e^{-2t}) && \text{multiply both sides by the denominator} \\ \\ 4 &= 1+9e^{-2t} && \text{divide both sides by 90} &&\\ \\ 3 &= 9e^{-2t} && \text{subtract 1 from both sides}&&\\ \\ 1/3 &= e^{-2t} && \text{divide both sides by 9}&& \end{align*}\]

Step 2:   Take the natural logarithm of both sides.

\[\begin{align*} \ln \left(\frac{1}{3} \right) &= \ln(e^{-2t}) \\ &= -2t && \hbox{cancellation property $\ln(e^x) = x$} \end{align*}\]

Step 3:   Solve for   \(t\).

\[\begin{align*} t &= -\frac{1}{2} \ln \left( \frac{1}{3}\right)\\ &= \frac{1}{2}\ln(3) && \hbox{since $\ln(1/3) = \ln(3^{-1}) = -\ln(3)$} \end{align*}\]

Example 4

Expand using properties of logarithms

Expand the following expression:

\[\ln \left(\frac{\sqrt[3]{(x+1)^2}\cdot e^{5x}}{x}\right).\]

Step 1:   Use the laws of logarithms to expand the given expression.

\[\begin{align*} \ln \left(\frac{\sqrt[3]{(x+1)^2}\cdot e^{5x}}{x}\right) &= \ln(\sqrt[3]{(x+1)^2}\cdot e^{5x}) - \ln(x) && \text{since $\ln(m/n) = \ln(m) - \ln(n)$} \\ \\ &= \ln(\sqrt[3]{(x+1)^2}) + \ln(e^{5x}) - \ln(x) && \text{since $\ln(mn) = \ln(m) + \ln(n)$}\\ \\ %&= \ln((x+1)^{2/3}) + \ln(e^{5x}) - \ln(x)\\ \\ &= \frac{2}{3} \ln(x+1) + 5x \ln(e) - \ln(x) && \text{sine $\ln(m^n) = n\ln(m)$}\\ \\ &= \frac{2}{3} \ln(x+1) + 5x - \ln(x) && \text{since $\ln(e) = 1$.} \end{align*}\]

Example 5

Equation of the tangent line

Find the tangent line to \(y=\dfrac{e^{27x}}{x^9}\) at the point \((1,e^{27})\).

Step 1:   Recall the point-slope equation of a line.

Point-Slope:

\[y = m(x-a) + b,\]

where \(m\) is the slope of the line and \((a,b)\) is a point on the line.

Step 2:   Compute the slope of the line by using the derivative.

Recall \(\dfrac{d}{dx}e^{f(x)} = e^{f(x)} f'(x)\).

\[y' = \frac{e^{27x}\cdot 27 \cdot x^9 - e^{27x}\cdot 9x^8}{x^{18}}.\]

Since the given point is \((1,e^{27})\), plug in \(x=1\) into the derivative to find the slope of the tangent line.

\[\begin{align*} m &= y'(1) \\ &= \frac{e^{27}\cdot 27 - e^{27}\cdot 9}{1}\\ &= 18e^{27}. \end{align*}\]

Step 3:   Write down the equation of the tangent line.

Since we were given the point \((1,e^{27})\) (i.e., \(a=1\) and \(b=e^{27}\)) and we found the slope (\(m=18e^{27}\)), we can now write down the equation of the tangent line using the point-slope equation of a line.

\[y = 18e^{27}(x-1) + e^{27}.\]

Example 6

Marginal revenue function

Suppose the unit selling price \(p(x)\) and the quantity supplied \(x\) of a certain product is given by

\[p(x) = x^3e^{5x}+12.\]

Find the marginal revenue function \(R'(x)\).

Step 1:   Find the revenue function,   \(R(x)\), using the formula   \(R(x) = x\cdot p(x)\).

\[R(x) = x^4e^{5x} +12x.\]

Step 2:   Compute the derivative of   \(R(x)\).

Recall \(\dfrac{d}{dx}e^{f(x)} = e^{f(x)} f'(x)\).

\[\begin{align*} R'(x) &= 4x^3(e^{5x})+x^4(e^{5x}\cdot 5) + 12 && \text{using product rule}\\ &= x^3 e^{5x}(4+5x) + 12. \end{align*}\]

Example 7

Derivative of a logarithmic function

Compute the derivative of \(f(x) = \ln\left(\dfrac{\sqrt{6x+1}}{5x}\right)\).

Step 1:   Expand   \(f(x)\)   using laws of logarithms.

\[\begin{align*} \ln\left(\dfrac{\sqrt{6x+1}}{5x}\right) &= \ln(\sqrt{6x+1}) - \ln(5x) && \hbox{since $\ln(m/n) = \ln(m) - \ln(n)$} \\ \\ &= \ln((6x+1)^{1/2}) - [\ln(5) + \ln(x)] && \hbox{since $\ln(mn) = \ln(m) + \ln(n)$} \\ \\ &= \frac{1}{2}\ln(6x+1) - \ln(5) - \ln(x) && \hbox{since $\ln(m^n) = n\ln(m).$} \end{align*}\]

Step 2:   Compute the derivative.

Recall \(\dfrac{d}{dx}\ln(f(x)) = \dfrac{1}{f(x)} f'(x)\).

\[\begin{align*} \frac{d}{dx}\left(\frac{1}{2}\ln(6x+1) - \ln(5) - \ln(x)\right) &= \frac{1}{2}\cdot \frac{1}{6x+1}\cdot 6 - 0 - \frac{1}{x} \\ &= \frac{3}{6x+1} - \frac{1}{x}. \end{align*}\]

Example 8

Implicit differentiation

Let \(\ln(xy)+y^7 = x^3 + 2x\). Find \(\dfrac{dy}{dx}\).

Step 1:   Differentiate both sides using implicit differentiation.

\[\frac{1}{xy} (y +xy') + 7y^6y' = 3x^2+2.\]

Step 2:   Multiply both sides by   \(xy\).

\[\begin{align*} xy\left[ \frac{1}{xy} (y +xy') + 7y^6y' \right] &= xy\frac{1}{xy} (y +xy') + xy\cdot 7y^6y' \\ &= (y+xy') + 7xy^7y'\\ \\ xy\left[ 3x^2 + 2 \right] &= 3x^3y + 2xy. \end{align*}\]

Therefore,

\[y+xy' + 7xy^7y' = 3x^3y + 2xy.\]

Step 3:   Rearrange terms.

Rearrange terms so that any term with a factor of   \(y'\)   is on the left-hand side of the equation and all other terms are on the right-hand side.

\[xy' + 7xy^7y' = 3x^3y + 2xy - y.\]

Step 4:   Factor out   \(y'\)   on the left-hand side.

\[y'(x + 7xy^7) = 3x^3y + 2xy - y.\]

Step 5:   Solve for   \(y'\).

\[y' = \frac{3x^3y + 2xy - y}{x + 7xy^7}.\]