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Exponential Functions - Problem Solving

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Complex numbers.

The beauty of Algebra through complex numbers, fractals, and Euler’s formula.

• Andrew Hayes

An exponential function is a function of the form \(f(x)=a \cdot b^x,\) where \(a\) and \(b\) are real numbers and \(b\) is positive. Exponential functions are used to model relationships with exponential growth or decay. Exponential growth occurs when a function's rate of change is proportional to the function's current value. Whenever an exponential function is decreasing, this is often referred to as exponential decay .

To solve problems on this page, you should be familiar with

• rules of exponents - algebraic
• solving exponential equations
• graphs of exponential functions .

Growth and Decay

Problem solving - basic, problem solving - intermediate, problem solving - advanced.

Suppose that the population of rabbits increases by 1.5 times a month. When the initial population is 100, what is the approximate integer population after a year? The population after \(n\) months is given by \(100 \times 1.5^n.\) Therefore, the approximate population after a year is \[100 \times 1.5^{12} \approx 100 \times 129.75 = 12975. \ _\square \]

Suppose that the population of rabbits increases by 1.5 times a month. At the end of a month, 10 rabbits immigrate in. When the initial population is 100, what is the approximate integer population after a year? Let \(p(n)\) be the population after \(n\) months. Then \[p(n+2) = 1.5 p(n+1) + 10\] and \[p(n+1) = 1.5 p(n) + 10,\] from which we have \[p(n+2) - p(n+1) = 1.5 \big(p(n+1) - p(n)\big).\] Then the population after \(n\) months is given by \[p(0) + \big(p(1) - p(0)\big) \frac{1.5^{n} - 1}{1.5 - 1} .\] Therefore, the population after a year is given by \[\begin{align} 100 + (160 - 100) \frac{1.5^{12} - 1}{1.5 - 1} \approx& 100 + 60 \times 257.493 \\ \approx& 15550. \ _\square \end{align}\]

Suppose that the annual interest is 3 %. When the initial balance is 1,000 dollars, how many years would it take to have 10,000 dollars? The balance after \(n\) years is given by \(1000 \times 1.03^n.\) To have the balance 10,000 dollars, we need \[\begin{align} 1000 \times 1.03^n \ge& 10000 \\ 1.03^n \ge& 10\\ n \log_{10}{1.03} \ge& 1 \\ n \ge& 77.898\dots. \end{align}\] Therefore, it would take 78 years. \( _\square \)

The half-life of carbon-14 is approximately 5730 years. Humans began agriculture approximately ten thousand years ago. If we had 1 kg of carbon-14 at that moment, how much carbon-14 in grams would we have now? The weight of carbon-14 after \(n\) years is given by \(1000 \times \left( \frac{1}{2} \right)^{\frac{n}{5730}}\) in grams. Therefore, the weight after 10000 years is given by \[1000 \times \left( \frac{1}{2} \right)^{\frac{10000}{5730}} \approx 1000 \times 0.298 = 298.\] Therefore, we would have approximately 298 g. \( _\square \)

Given three numbers such that \( 0 < a < b < c < 1\), define

\[ A = a^{a}b^{b}c^{c}, \quad B = a^{a}b^{c}c^{b} , \quad C = a^{b}b^{c}c^{a}. \]

How do the values of \(A, B, C \) compare to each other?

\[\large 2^{x} = 3^{y} = 12^{z} \]

If the equation above is fulfilled for non-zero values of \(x,y,z,\) find the value of \(\frac { z(x+2y) }{ xy }\).

If \(5^x = 6^y = 30^7\), then what is the value of \( \frac{ xy}{x+y} \)?

If \(27^{x} = 64^{y} = 125^{z} = 60\), find the value of \(\large\frac{2013xyz}{xy+yz+xz}\).

\[\large f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}} \]

Suppose we define the function \(f(x) \) as above. If \(f(a)=\frac{5}{3}\) and \(f(b)=\frac{7}{5},\) what is the value of \(f(a+b)?\)

\[\large \left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{2000}\right)^{2000}\]

Given that \(x\) is an integer that satisfies the equation above, find the value of \(x\).

\[\Large a^{(a-1)^{(a-2)}} = a^{a^2-3a+2}\]

Find the sum of all positive integers \(a\) that satisfy the equation above.

Find the sum of all solutions to the equation

\[ \large (x^2+5x+5)^{x^2-10x+21}=1 .\]

\[\large |x|^{(x^2-x-2)} < 1 \]

If the solution to the inequality above is \(x\in (A,B) \), then find the value of \(A+B\).

Master concepts like these

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Precalculus : Exponential Equations and Inequalities

Study concepts, example questions & explanations for precalculus, all precalculus resources, example questions, example question #1 : use logarithms to solve exponential equations and inequalities.

Solving an exponential equation.

We recall the property:

Example Question #132 : Exponential And Logarithmic Functions

Example Question #3 : Use Logarithms To Solve Exponential Equations And Inequalities

Expanding the logarithms into sums of logarithms will cancel out the first two x terms, resulting in the equation:

Combining the first and second terms, then subtracting the new term over will allow you to isolate the variable term.

Divide both sides of the equation by 2, then exponentiate with 3.

Evaluating this term numerically will give the correct answer.

Example Question #4 : Use Logarithms To Solve Exponential Equations And Inequalities

Solve the following equation:

To solve this equation, recall the following property:

Evaluate with your calculator to get

Example Question #141 : Exponential And Logarithmic Functions

After using the division rule to simplify the left hand side you can take the natural log of both sides.

If you then combine like terms you get a quadratic equation which factors to,

Example Question #6 : Use Logarithms To Solve Exponential Equations And Inequalities

Example Question #1 : Exponential Equations And Inequalities

Example Question #9 : Use Logarithms To Solve Exponential Equations And Inequalities

Example Question #10 : Use Logarithms To Solve Exponential Equations And Inequalities

Cannot be simplified any further

The logarithmic expression is as simplified as can be. 

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General Mathematics Quarter 1 – Module 18: Solving Exponential Equations and Inequalities

This module was designed and written with you in mind. It is here to help you master how to solve exponential equation and inequality. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

After going through this module, you are expected to:

1. explain how to apply the properties in solving exponential equations and inequalities; and

2. solve exponential equations and inequalities.

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6.4: Logarithmic Equations and Inequalities

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• Page ID 119171

• Carl Stitz & Jeff Zeager
• Lakeland Community College & Lorain County Community College

Math 370 Learning Objectives

• Review solving logarithmic equations.
• Optional: Solve logarithmic inequalities.
• Optional: Find the inverse when given an equation involving logarithms.

Solving Logarithmic Equations

In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve \(\log_{2}(x) = \log_{2}(5)\). Theorem 6.2.2  tells us that the only solution to this equation is \(x=5\). Now suppose we wish to solve \(\log_{2}(x) = 3\). If we want to use Theorem 6.2.2 , we need to rewrite \(3\) as a logarithm base \(2\). We can use Theorem 6.2.1  to do just that: \(3 = \log_{2}\left(2^{3}\right) = \log_{2}(8)\). Our equation then becomes \(\log_{2}(x) = \log_{2}(8)\) so that \(x = 8\). However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.2.1 to rewrite the equation \(\log_{2}(x) = 3\) as \(2^{3} = x\), or \(x=8\).

Example \( \PageIndex{1} \)

• \(\log_{117}(1-3x) = \log_{117}\left(x^2-3\right)\)
• \(2 - \ln(x-3) = 1\)
• \(\log_{6}(x+4) + \log_{6}(3-x) = 1\)
• \(\log_{7}(1-2x) = 1 - \log_{7}(3-x)\)
• \(\log_{2}(x+3) = \log_{2}(6-x)+3\)
• \(1 + 2 \log_{4}(x+1) = 2 \log_{2}(x)\)
• Since we have the same base on both sides of the equation \(\log_{117}(1-3x) = \log_{117}\left(x^2-3\right)\), we equate what’s inside the logs to get \(1-3x = x^2-3\). Solving \(x^2+3x-4 = 0\) gives \(x=-4\) and \(x=1\). To check these answers using the calculator, we make use of the Change of Base Formula and graph \(f(x) = \frac{\ln(1-3x)}{\ln(117)}\) and \(g(x) = \frac{\ln\left(x^2-3\right)}{\ln(117)}\) and we see they intersect only at \(x=-4\). To see what happened to the solution \(x=1\), we substitute it into our original equation to obtain \(\log_{117}(-2) = \log_{117}(-2)\). While these expressions look identical, neither is a real number, 1  which means \(x=1\) is not in the domain of the original equation, and is not a solution.

• Taking a cue from the previous problem, we begin solving \(\log_{7}(1-2x) = 1 - \log_{7}(3-x)\) by first collecting the logarithms on the same side, \(\log_{7}(1-2x) + \log_{7}(3-x) = 1\), and then using the Product Rule for Logarithmic Functions to get \(\log_{7}[(1-2x)(3-x)] = 1\). Rewriting this as an exponential equation gives \(7^{1} = (1-2x)(3-x)\) which gives the quadratic equation \(2x^2-7x-4=0\). Solving, we find \(x = -\frac{1}{2}\) and \(x=4\). Graphing, we find \(y = f(x) = \frac{\ln(1-2x)}{\ln(7)}\) and \(y=g(x) = 1 - \frac{\ln(3-x)}{\ln(7)}\) intersect only at \(x=-\frac{1}{2}\). Checking \(x=4\) in the original equation produces \(\log_{7}(-7) = 1 - \log_{7}(-1)\), which is a clear domain violation.

If nothing else, Example \( \PageIndex{1} \) demonstrates the importance of checking for extraneous solutions 2  when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1 , much can be learned from checking all of the answers in Example \( \PageIndex{1} \) analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions.

Subsection Footnotes

1  They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables.

2  Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.

Solving Logarithmic Inequalities

Since logarithmic functions are continuous on their domains, we can use sign diagrams.

Example \( \PageIndex{2} \)

Solve the following inequalities. Check your answer graphically using a calculator.

• \(\frac{1}{\ln(x)+1} \leq 1\)
• \(\left(\log_{2}(x)\right)^2 < 2 \log_{2}(x) + 3\)
• \(x \log(x+1) \geq x\)

​​​​​​Our next example revisits the concept of pH first seen in the homework exercises of Section 6.1 .

Example \( \PageIndex{3} \)

In order to successfully breed Ippizuti fish the pH of a freshwater tank must be at least \(7.8\) but can be no more than \(8.5\). Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator.

From the homework exercises in  Section 6.1 , \(\text{pH} = -\log[\text{H}^{+}]\) where \([\text{H}^{+}]\) is the hydrogen ion concentration in moles per liter. We require \(7.8 \leq -\log[\text{H}^{+}] \leq 8.5\) or \(-7.8 \geq \log[\text{H}^{+}] \geq -8.5\). To solve this compound inequality we solve \(-7.8 \geq \log[\text{H}^{+}]\) and \(\log[\text{H}^{+}] \geq -8.5\) and take the intersection of the solution sets. 

The former inequality yields \(0 < [\text{H}^{+}] \leq 10^{-7.8}\) and the latter yields \([\text{H}^{+}] \geq 10^{-8.5}\). Taking the intersection gives us our final answer \(10^{-8.5} \leq [\text{H}^{+}] \leq 10^{-7.8}\). (Your Chemistry professor may want the answer written as \(3.16 \times 10^{-9} \leq [\text{H}^{+}] \leq 1.58 \times 10^{-8}\).) After carefully adjusting the viewing window on the graphing calculator we see that the graph of \(f(x) = -\log(x)\) lies between the lines \(y = 7.8\) and \(y = 8.5\) on the interval \([3.16 \times 10^{-9}, 1.58 \times 10^{-8}]\).

Finding the Inverse When Given an Equation Involving Logarithms

We close this section by finding an inverse of a one-to-one function which involves logarithms.

Example \( \PageIndex{4} \)

The function \(f(x) = \frac{\log(x)}{1-\log(x)}\) is one-to-one. Find a formula for \(f^{-1}(x)\) and check your answer graphically using your calculator.

We first write \(y=f(x)\) then interchange the \(x\) and \(y\) and solve for \(y\).

\[\begin{array}{rclr} y & = & f(x) & \\ y & = & \dfrac{\log(x)}{1-\log(x)} & \\ x & = & \dfrac{\log(y)}{1-\log(y)} & \left(\text{Interchange }x\text{ and }y\right)\\ x\left(1-\log(y)\right) & = & \log(y) & \\ x - x\log(y) & = & \log(y) & \\ x & = & x \log(y) + \log(y) & \\ x & = & (x+1) \log(y) & \\ \dfrac{x}{x+1} & = & \log(y) & \\ y & = & 10^{\frac{x}{x+1}} & \left(\text{Rewrite as an exponential equation}\right) \\ \end{array}\nonumber\]

We have \(f^{-1}(x) = 10^{\frac{x}{x+1}}\). Graphing \(f\) and \(f^{-1}\) on the same viewing window yields

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Name: Mr. Victor Lee Class Days: Monday, Wednesday Class Times: 2:15 pm – 3:55 pm Room: N703 Office: N825 Office Hours: Monday, Wednesday 11:00 am – 12:00 pm Online Classroom/Office

Textbook: Precalculus ( 3rd Edition) by Tradler and Carley

Topics include an in-depth study of functions such as polynomial functions, radical functions, trigonometric functions, exponential and logarithmic functions; connections to vectors and complex numbers; solving trigonometric equations, and identities involving sum, double and half-angle formulas; Binomial Theorem and progressions.  

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