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#

  • For all \(x>0\)

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

  • For all \(x\)

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

Derivatives#

  • 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#

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

Example 2#

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#

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

Example 5#

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:

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

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.

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

Example 6#

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#

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#

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