Honors Pre-Calculus A, 3rd Hour, Winter 2016: 3.4 Solving Exponential and Logarithmic Equations

Exponential and logarithmic equations can be solved using two strategies based off of the One-to-One Properties and Inverse Properties.

For $a > 0$ and $a \neq 1$ , these properties are true for all $x$ and $y$ for which $log_{a}x$ and $log_{a}y$ are defined.

One-to-One Properties:

1). $a^{x}=a^{y}$ if and only if $x=y$

ex: $5^{x}=125$

$5^{x}=5^{3}$

$x=3$

2). $\mathrm{log}_{a}x= \mathrm{log}_{a}y$ if and only if $x=y$

ex:

$\mathrm{log}_{2}4-\mathrm{log}_{2}x=0$

$\mathrm{log}_{2}4=\mathrm{log}_{2}x$

$x=4$

Inverse Properties:

1). $a^{\mathrm{log_{a}}}x=x$

ex:

$\mathrm{log}x=2$

$10^{\mathrm{log}x}=10^{2}$

$x=10^{2}$

$x=100$

2). $\mathrm{log}_{a}a^{x}=x$

ex:

$\mathrm{log}_{7}7^{x}=4$

$x= 4$

The examples above are the very basic to clearly show how the one-to-one and inverse properties can be used to solve logarithmic and exponential equations. The strategies to solve logarithmic and exponential equations can be summarized through the following:

Rewrite the given equation in a form to use the One-to-One properties of exponential or logarithmic functions.
Rewrite an exponential equation in logarithmic form and apply the inverse property of logarithmic functions.
Rewrite a logarithmic equation in exponential form and apply the Inverse property of exponential functions.

Change-of-Base formula

The Change-of-Base formula shows that: $\mathrm{log}_{a}x=\frac{\mathrm{log_{b}}x}{\mathrm{log}_{b}a}$

This is proved through the following:

let $y= \mathrm{log}_{a}x$

$\mathrm{log}_{b}a^{y}=\mathrm{log}_{b}x$

$y\mathrm{log}_{b}a=\mathrm{log}_{b}x$

$y= \frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$

$\mathrm{log}_{a}x= \frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$

Solving Exponential Equations

Algebraic Solution:

Solve: $8e^{3x}=10$

$e^{3x}=\frac{5}{4}$ divide both sides by 8 and simplify the fraction

$\mathrm{ln} e^{3x}=\mathrm{ln}\frac{5}{4}$ Take logarithm of each side

${3x}=\mathrm{ln}\frac{5}{4}$ Use the inverse property

$x=\frac{1}{3}\mathrm{ln}\frac{5}{4}$ solve for x by multiplying each side by 1/3

Graphical Solution:

Enter $y_{1}=8e^{^{3}x}$ and $y_{2}=10$ into your calculator

use the intersect or zoom and trace features to approximate the intersection point. This graph shows that the two equations intersect at (0.074,10) or $(\frac{1}{3}ln\frac{5}{4},10)$

Graphical Solution 2:

rewrite the original equation as $y_{1}= 8e^{3x}-10$

Use the zero root feature or the zoom and trace features to approximate the x-intercept(s). the zero on this graph is shown at (0.074,0) or $(\frac{1}{3}ln\frac{5}{4},0)$

Solving an Exponential Equation in Quadratic Form:

Algebraic solution:

Solve: $e^{2x}-5e^{x}+4=0$

$(e^{x})^{2}-5e^{x}+4$ Rewrite in quadratic form

$(e^{x}-4)(e^{x}-1)$ Factor

$(e^{x}-4)=0$ $(e^{x}-1)=0$ Set both factors equal to zero

$e^{x}=4$ $e^{x}=1$ isolate ${\color{Red} e^{x}}$

$x= \mathrm{ln}4$ $x= \mathrm{ln}1$

*Check both of these solutions in the original equation to see if either is extraneous*

To find the solutions to this equation graphically, use the second graphical solution method shown above.

Solving Logarithmic Equation

Algebraic Solution:

Solve: $6+ 2\mathrm{log}_{2}x=5$

$2\mathrm{log}_{2}x=-1$ subtract 6 from each side

$\mathrm{log}_{2}x=\frac{-1}{2}$ divide each side by 2

$2^{\mathrm{log}_{2}x}=2^{\frac{-1}{2}}$ exponentiate each side

$x= 2^{\frac{-1}{2}}$ Inverse property to solve

$x=\frac{\sqrt{2}}{2}$

Graphical Solution:

Enter $y_{1}= 6+ 2\mathrm{log}_{2}x$ and $y_{1}= 5$ into your graphing utility

use the intersect or zoom and trace features to approximate the intersection point. This graph shows that the two equations intersect at (0.707, 5). 0.707= $\frac{\sqrt{2}}{2}$

Checking for Extraneous Solutions:

Algebraic Equation:

solve: $\mathrm{ln}(x-2)+\mathrm{ln}(2x-3)=2\mathrm{ln}x$

$\mathrm{ln}\left [ (x-2)\left ( 2x-3 \right ) \right ]=lnx^{2}$ Use the properties of logarithms

$2x^{2}-7x+6=x^{2}$ One-to-One Property

$x^{2}-7x+6=0$ Write in general form

$\left ( x-6 \right )(x-1)=0$ Factor

$x=6$ $x=1$

By checking both of the solutions with the original equation, we can conclude that x=1 is an extraneous answer. This is because the solution x=1 creates the equation $\mathrm{ln}(-1)+\mathrm{ln}(-1)=2\mathrm{ln}1$ , and -1 is outside of the domain of a natural log function.

Graphical Solution:

Rewrite the original equation as $y_{1}=\mathrm{ln}\left ( x-2 \right )+\mathrm{ln}(2x-3)-2\mathrm{ln}$

Use the zero root feature or the zoom and trace features to approximate the x-intercept(s). the zero on this graph is shown at (6,0).

Approximating Solutions

This method should be utilized for equations that contain combinations of algebraic functions, exponential functions, and/or logarithmic functions that will be messy by solving algebraically.

example:

solve: $\mathrm{ln}x=x^{2}-2$