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2.13 Exponential and logarithmic equations and inequalities

Lesson

Introduction

Learning objectives

  • 2.13.A Solve exponential and logarithmic equations and inequalities.
  • 2.13.B Construct the inverse function for exponential and logarithmic functions.

Solving exponential equations

Certain exponential equations arise that can be easily solved without resorting to logarithms. The key to their solution lies in the recognition of various powers of integers. We can solve these types of equations by rewriting both sides of the equation with the same base to some power. If we can do this, we can use the equality property of exponents which says b^x=b^y \iff x=y If both sides of the equation cannot be written in the same base, another method of solving exponential equations is taking the logarithm of both sides. Then, we can use properties of logarithms to solve the equation. Often, we will evaluate the solutions to these equations with a calculator.

If these methods do not work, we can use technology to estimate solutions. To solve with a graphing calculator, we can graph the expression on the left side of the equation, graph the expression on the right side of the equation, and then find their point of intersection.

Examples

Example 1

Solve each equation for x.

a

2^{1-2x}=\dfrac{1}{512}

Worked Solution
Create a strategy

The denominator of the right side of the equation, 512, is a power of 2. By first rewriting this as a power with a base of 2, we can simplify the process needed to solve this equation.

Apply the idea
\displaystyle 2^{1-2x}\displaystyle =\displaystyle \frac{1}{512}State the equation
\displaystyle 2^{1-2x}\displaystyle =\displaystyle \frac{1}{2^{9}}Rewrite 512 as a power of 2
\displaystyle 2^{1-2x}\displaystyle =\displaystyle 2^{-9}Negative exponent property
\displaystyle 1-2x\displaystyle =\displaystyle -9Equality property of exponents
\displaystyle -2x\displaystyle =\displaystyle -10Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle 5Division property of equality
Reflect and check

In general, a fraction in an exponential equation will typically lead to a negative exponent.

b

e^{2x+3}=4

Worked Solution
Create a strategy

We cannot rewrite 4 as e to some power, so we cannot use that method to solve this equation. The inverse of e^x is \ln\left(x\right), so we can use the equality property of logarithms to take the natural logarithm of both sides. Then, use the inverse property of logarithms to simplify.

Apply the idea
\displaystyle e^{2x+3}\displaystyle =\displaystyle 4Original equation
\displaystyle \ln(e^{2x+3})\displaystyle =\displaystyle \ln(4)Equality property of logarithms
\displaystyle 2x+3\displaystyle =\displaystyle \ln(4)Inverse property of natural logarithms
\displaystyle 2x\displaystyle =\displaystyle \ln(4)-3Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle \frac{\ln(4)-3}{2}Division property of equality

The exact answer is x=\dfrac{\ln\left(4\right)-3}{2}.

Reflect and check

We can find an approximate solution using technology, x\approx -0.8069.

c

2^{3x-1}=5^{x}

Worked Solution
Create a strategy

We cannot rewrite 2 and 5 as powers with the same base, and taking the logarithm of both sides would be complicated. Estimating the solution using technology is the simplest method for solving this equation.

Apply the idea

First, we need to graph y=2^{3x-1} and y=5^x on the same plane.

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Then, we need to adjust the window to view the intersection of the lines. The lines appear to intersect between x=1 and x=2 and between y=9 and y=12.

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We can use the intersect feature of a graphing calculator to find that the lines intersect at x\approx 1.475.

Reflect and check

Solving this equation by taking the logarithm of both sides is complicated but possible.

\displaystyle 2^{3x-1}\displaystyle =\displaystyle 5^xOriginal equation
\displaystyle \log 2^{3x-1}\displaystyle =\displaystyle \log 5^xEquality property of logarithms
\displaystyle \left(3x-1\right)\log 2\displaystyle =\displaystyle x\log 5Power property of logarithms
\displaystyle 3x\log 2-\log 2\displaystyle =\displaystyle x\log 5Distributive property
\displaystyle 3x\log 2-\log 2-x\log 5\displaystyle =\displaystyle 0Subtraction property of equality
\displaystyle 3x\log 2-x\log 5\displaystyle =\displaystyle \log 2Addition property of equality
\displaystyle x\left(3\log 2-\log 5\right)\displaystyle =\displaystyle \log 2Distributive property
\displaystyle x\displaystyle =\displaystyle \frac{\log 2}{3\log 2-\log 5}Division property of equality

If we used a calculator to approximate the decimal, we would find x\approx1.475.

Example 2

A video posted online initially had 4 views as soon as it was posted. The total number of views to date has been increasing by approximately 12\% each day. Determine when the video will reach 1 million views.

Worked Solution
Create a strategy

We can write an exponential equation to model this situation. The exponential form of an equation is f\left(x\right)=ab^x where a is the initial value, b is the growth factor, x is the number of days, and f\left(x\right) is the number of views.

The number of views increases by 12\%, so the growth factor is 1+0.12=1.12.

Apply the idea

The equation that models the number of views is f\left(x\right)=4\left(1.12\right)^x.

We can solve for when the video reaches 1 million views by setting f\left(x\right) equal to 1\,000\,000.

\displaystyle 4\left(1.12\right)^x\displaystyle =\displaystyle 1\,000\,000Set f\left(x\right) equal to 1 million views
\displaystyle \left(1.12\right)^x\displaystyle =\displaystyle 250\,000Division property of equality
\displaystyle x\displaystyle =\displaystyle \log_{1.12}\left(250\,000\right)Rewrite as a logarithmic equation
\displaystyle x\displaystyle =\displaystyle \frac{\log{\left(250\,000\right)}}{\log\left(1.12\right)}Change of base property
\displaystyle x\displaystyle \approx\displaystyle 109.67405Evaluate using technology

Assuming the views continue to increase by 12\% per day, the video will reach 1 million views in 110 days.

Reflect and check

Another way to solve this is by using the equality property of logarithms after the second step:

\displaystyle 4\left(1.12\right)^x\displaystyle =\displaystyle 1\,000\,000Set f\left(x\right) equal to 1 million views
\displaystyle \left(1.12\right)^x\displaystyle =\displaystyle 250\,000Division property of equality
\displaystyle \log\left(1.12\right)^x\displaystyle =\displaystyle \log\left(250\,000\right)Equality property of logarithms
\displaystyle x\log\left(1.12\right)\displaystyle =\displaystyle \log\left(250\,000\right)Power property of logarithms
\displaystyle x\displaystyle =\displaystyle \frac{\log\left(250\,000\right)}{\log\left(1.12\right)}Division property of equality
\displaystyle x\displaystyle \approx\displaystyle 109.67405Evaluate using technology
Idea summary

There are three main strategies for solving exponential equations:

  • Rewrite both sides of the equation with the same base to some power

  • Rewrite in logarithmic form or use the equality property to introduce logarithms on either side

  • If neither of the above strategies works, estimate the solution(s) using technology

Solving logarithmic equations

Many logarithmic equations can be solved by rewriting it in exponential form. The properties of logarithms may need to be used to condense the equation down to one logarithm before switching it to exponential form.

Recall a logarithmic function of the form \log_b\left(x\right) has domain x>0. When solving logarithmic equations, we need to ensure our answers make sense in context and that we are never taking the logarithm of a negative number. We may find that, for some logarithmic equations, a solution resulting from the process of solving is extraneous because it results in a negative argument.

Logarithmic equations which have non-zero expressions on both sides can be solved graphically. To do this, set the expressions on both sides equal to y, graph each equation on the same coordinate plane, then find the point(s) of intersection.

Examples

Example 3

Solve each equation, indicating whether each solution is viable or extraneous.

a

\dfrac{1}{3}\log_{5} x +7=8

Worked Solution
Create a strategy

The first thing we want to do is isolate the logarithm. Then, we can rewrite it in exponential form and solve for x. The domain of the logarithm is x>0.

Apply the idea
\displaystyle \frac{1}{3}\log_{5} x +7\displaystyle =\displaystyle 8Original equation
\displaystyle \frac{1}{3}\log_{5} x\displaystyle =\displaystyle 1Subtraction property of equality
\displaystyle \log_{5} x\displaystyle =\displaystyle 3Multiplication property of equality
\displaystyle x\displaystyle =\displaystyle 5^3Definition of logarithm
\displaystyle x\displaystyle =\displaystyle 125Evaluate the exponent

The result lies within the domain, so the solution x=125 is valid.

Reflect and check

When we get to the step \log_5 x =3, we can use the equality property of exponents instead of switching it to exponential form.

\displaystyle \log_5 x\displaystyle =\displaystyle 3
\displaystyle 5^{\log_5 x}\displaystyle =\displaystyle 5^3
\displaystyle x\displaystyle =\displaystyle 5^3
\displaystyle x\displaystyle =\displaystyle 125

This method is helpful when solving logarithmic inequalities.

b

\log_6 \left(x\right)+\log_6 \left(x+9\right)=2

Worked Solution
Create a strategy

We can rewrite the left hand side using the product property of logarithms: \log_b\left(x\right)+ \log_b\left(y\right) =\log_b\left(xy\right) We can then use the identity \log_b x=n \iff x=b^n , to rewrite the expression without logarithms.

Apply the idea
\displaystyle \log_6 x+\log_6 \left(x+9\right)\displaystyle =\displaystyle 2State the equation
\displaystyle \log_6 \left[x\left(x+9\right)\right]\displaystyle =\displaystyle 2Product property of logarithms
\displaystyle \log_6\left(x^2+9x\right)\displaystyle =\displaystyle 2Distributive property
\displaystyle x^2+9x\displaystyle =\displaystyle 6^2Definition of logarithm
\displaystyle x^2+9x\displaystyle =\displaystyle 36Evaluate the square
\displaystyle x^2+9x-36\displaystyle =\displaystyle 0Subtraction property of equality
\displaystyle \left(x-3\right)\left(x+12\right)\displaystyle =\displaystyle 0Factor the quadratic

This gives us two possible solutions x=3 and x=-12.

To test for extraneous solutions, we want to check if any solution falls outside the domain of the logarithms. The logarithms in the original equation are \log_6 x and \log_6 \left(x+9\right), and we cannot take the logarithm of a negative value.

According to the first term, x must be greater than 0. According to the second term, x must be greater than -9. However, any number between -9 and 0 will make the first term undefined, so the domain for this equation is \left(0, \infty\right).

We now need to check if either solution is extraneous. As -12 < 0, it falls outside the domain of the function and is therefore extraneous.

x=3 is a valid solution.

Reflect and check

We can check our solution by substituting x=3 into the equation:

\displaystyle \log_6\left(3\right)+\log_6\left(3+9\right)\displaystyle =\displaystyle \log_6\left(3\cdot 12\right)
\displaystyle =\displaystyle \log_6\left(36\right)
\displaystyle =\displaystyle \log_6\left(6^2\right)
\displaystyle =\displaystyle 2

Substituting x=3 makes the equation true, so we confirm that it is a valid solution.

c

\log _n\left(x+4\right)-\log _n\left(x-2\right)=\log _n\left(x\right)

Worked Solution
Create a strategy

We can rewrite the left hand side using the quotient property of logarithms: \log_b\left(x\right)- \log_b\left(y\right) =\log_b\left(\frac{x}{y}\right) This will give us an equation of the form \log_b\left(x\right)=\log_b\left(y\right) which we can simplify using the equality property\log_b\left(x\right)=\log_b\left(y\right) \iff x=y

Apply the idea
\displaystyle \log _n\left(x+4\right)-\log _n\left(x-2\right) \displaystyle =\displaystyle \log _n\left(x\right)State the equation
\displaystyle \log _n\left(\frac{x+4}{x-2}\right)\displaystyle =\displaystyle \log _n\left(x\right)Quotient property of logarithms
\displaystyle \frac{x+4}{x-2}\displaystyle =\displaystyle xEquality property of logarithms
\displaystyle x+4\displaystyle =\displaystyle x\left(x-2\right)Multiplication property of equality
\displaystyle x+4\displaystyle =\displaystyle x^2-2xDistributive property
\displaystyle 0\displaystyle =\displaystyle x^2-3x-4Subtraction property of equality
\displaystyle 0\displaystyle =\displaystyle \left(x+1\right)\left(x-4\right)Factor the quadratic

This gives us two possible solutions x=-1 and x=4.

To test for extraneous solutions, we want to check if any solution falls outside the domain of the logarithms. The logarithmic terms in the original equation are \log_n\left(x+4\right), \log_n\left(x-2\right), and \log_n\left(x\right).

LogarithmDomain
\log_n \left(x+4\right)\left(-4,\infty\right)
\log_n \left(x-2\right)\left(2,\infty\right)
\log_n x\left(0,\infty\right)

Any number between -4 and 2 will make the second term undefined, so the domain for this equation is \left(2, \infty\right).

We now need to check if either solution is extraneous. As -1 <2, it falls outside the domain of the function and is therefore extraneous.

x=4 is a valid solution.

Reflect and check

We can check our solution by substituting x=4 into the equation:

\displaystyle \log_n\left(x+4\right)-\log_n\left(x-2\right)\displaystyle =\displaystyle \log_n\left(x\right)
\displaystyle \log_n\left(4+4\right)-\log_n\left(4-2\right)\displaystyle =\displaystyle \log_n\left(4\right)
\displaystyle \log_n\left(8\right)-\log_n\left(2\right)\displaystyle =\displaystyle \log_n\left(4\right)
\displaystyle \log_n\left(\frac{8}{2}\right)\displaystyle =\displaystyle \log_n\left(4\right)
\displaystyle \log_n\left(4\right)\displaystyle =\displaystyle \log_n\left(4\right)

Example 4

Consider the equation 7\log_{10}\left(x+3\right)=x+2.

a

Sketch the graph of y=7\log_{10}\left(x+3\right).

Worked Solution
Create a strategy

We can graph this equation by first completing a table of values and drawing the curve that passes through these points, or we can consider how the function f\left(x\right)=\log_{10}\left(x\right) has been transformed.

Considering each component separately, we can see that the graph of y=\log_{10}\left(x\right) is stretched vertically by a factor of 7 and translated 3 units to the left.

Apply the idea
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  • First, the graph of y=\log_{10}\left(x\right) is stretched vertically by a factor of 7 .
  • This corresponds with the graph of {y=7\log_{10}\left(x\right)}.
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  • Next, the graph of f\left(x\right)=7\log_{10}\left(x\right) is translated 3 units to the left.
  • This corresponds with the graph of {y=7\log_{10}\left(x+3\right)}.
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Reflect and check

Remember that we apply stretches, compressions, and reflections before applying translations.

b

Graph y=x+2 on the same coordinate plane.

Worked Solution
Create a strategy

The graph of y=x+2 is a straight line with a slope of 1 and a y-intercept of 2.

Apply the idea
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c

Estimate the solutions to the equation, rounding to the nearest integer.

Worked Solution
Create a strategy

The solutions to the equation will be the points where the two graphs intersect.

Apply the idea
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Rounding to the nearest integer, we get the solutions x=-2 and x=4.

Reflect and check

We can confirm that x=-2 is exactly a solution algebraically, as the left hand side of the equation becomes 7\log_{10}\left(1\right)= 0, and the right hand side becomes -2+2=0, as expected. We cannot do the same for x=4, which tells us that this solution must have been rounded.

Idea summary

To solve logarithmic equations, we can use properties of logarithms to condense the terms and then use the definition of logarithms to switch to exponential form. If this is not possible or practical, we can estimate the solution(s) using technology.

Solving exponential inequalities

Exponential inequalities are inequalities that include an exponent. These inequalities come in various forms, such as b^{x} \gt k,\, b^{x} \lt k,\, b^{x} \geq k, or b^{x}\leq k, where b is the base, x is the exponent, and k is a constant. The base, b, is a positive number not equal to 1. Just like exponential equations, in an exponential inequality, the base of the exponential expression can be a number, a variable, or even a more complicated expression. It's also important to note that exponential inequalities, like exponential equations, deal with exponential growth or decay situations.

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Solving exponential inequalities often requires a good understanding of both exponential functions and logarithmic functions. That's because logarithms are the inverses of exponential functions, and they provide us with the tools we need to solve for variables that are in the exponent of an exponential inequality.

To solve an exponential inequality using logarithms, we generally start by isolating the exponential expression on one side of the inequality. Then, we can apply a logarithm to both sides of the inequality. This allows us to "bring down" the exponent and turn it into a coefficient, which is often easier to work with.

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The graph above shows an exponential function, y = 2^{x}, and its corresponding logarithmic function, y = \text{log}_{2}x. You'll notice that these functions are reflections of each other across the line y = x. This illustrates the idea that logarithms "undo" or "reverse" the operation of exponentiation, which is crucial for understanding how to solve exponential inequalities.

Remember, when dealing with inequalities, it's important to keep in mind that the direction of the inequality sign may change when both sides are multiplied or divided by a negative number or when taking the logarithm of both sides, as \text{log}_{b}a \gt \text{log}_{b}c if and only if a \gt c for b \gt 1.

Examples

Example 5

Solve the inequality 4.5^{x} \gt 529.

Worked Solution
Create a strategy

We can start by isolating the exponential expression on one side of the inequality. Then we can apply a logarithm to both sides to "bring down" the exponent, which makes it easier to solve for the variable x.

Apply the idea

Let's find the critical points of the inequality.

\displaystyle 4.5^{x}\displaystyle =\displaystyle 529
\displaystyle \text{log}(4.5^{x})\displaystyle =\displaystyle \text{log}(529)Take log of both sides
\displaystyle x\times \text{log}(4.5)\displaystyle =\displaystyle \text{log}(529)Move the exponent x out in front as a multiplier
\displaystyle x\displaystyle =\displaystyle \dfrac{\text{log}(529)}{\text{log}(4.5)}Divide both sides by \text{log}(4.5)
\displaystyle x\displaystyle \approx\displaystyle 4.2Evaluate

So, the answer is x \gt 4.2.

Reflect and check

To check the solution, we can substitute the x = 4.2 to the original inequality.

\displaystyle 4.5^{4.2}\displaystyle >\displaystyle 529Substitute x = 4.2
\displaystyle 554\displaystyle >\displaystyle 529Evaluate

We have 4.5^{4.2} which is indeed greater than 529, so our solution is correct.

Idea summary

Exponential inequalities are similar to exponential equations, but they involve an inequality sign (\gt or \lt) instead of an equals sign. We solve exponential inequalities by applying our knowledge of exponents and inequalities.

Solving logarithmic inequalities

Much like exponential inequalities, logarithmic inequalities involve a comparison of two logarithmic expressions or a logarithmic expression and a number. The main difference is that, in this case, the variable we're solving for is in the argument of the logarithm, not the exponent. For example, an inequality like \text{log}_\text{base} b (x) \gt a.

Remember the important properties of logarithms that you've learned before. The logarithm is the inverse operation of exponentiation, which means that they undo each other. Also, logarithms have a key property that \text{log} (a \times b) = \text{log}(a) + \text{log}(b), and \text{log}\left(\dfrac{a}{b}\right) = \text{log}(a) - \text{log}(b). These properties are helpful when solving logarithmic inequalities.

Solving logarithmic inequalities generally involves using properties of logarithms, converting to exponential form, and careful consideration of the domain of the logarithm.

Remember that the domain of a logarithm is all positive real numbers. That is, we can only take the logarithm of a positive number. This is crucial to remember when solving logarithmic inequalities because it affects the solutions of the inequality.

Moreover, when solving inequalities involving logarithms, the direction of the inequality symbol depends on the base of the logarithm:

  • If the base is greater than 1 (such as in \text{log}_2(x) or \text{log}_10(x)), the inequality symbol remains the same when you convert the inequality from logarithmic form to exponential form.
  • If the base is between 0 and 1 (like in \text{log}_{0.5}(x) or \text{log}_\frac{1}{3}(x)), the inequality symbol flips direction when you convert the inequality from logarithmic form to exponential form.

Examples

Example 6

Solve the inequality \text{log}(-4x+6)\geq 8.

Worked Solution
Create a strategy

Convert the logarithmic inequality into an exponential form, and then isolate the x term and find the value of x that satisfies the inequality.

Apply the idea
\displaystyle \text{log}(-4x+6)\displaystyle \geq\displaystyle 8Write the inequality
\displaystyle -4x+6\displaystyle \geq\displaystyle 10^{8}Convert to exponential form
\displaystyle -4x+6\displaystyle \geq\displaystyle 100\,000\,000Simplify the right side
\displaystyle -4x\displaystyle \geq\displaystyle 100\,000\,000 - 6Isolate the x term
\displaystyle -4x\displaystyle \geq\displaystyle 99\,999\,994Simplify the right side
\displaystyle x\displaystyle \geq\displaystyle -\dfrac{99\,999\,994}{4}Divide both sides by -4
\displaystyle x\displaystyle \leq\displaystyle -24\,999\,998.5Evaluate

Remember, when you divide or multiply by a negative number, the direction of the inequality flips.

Our solution is x \leq -24,999,998.5. This means that any number less than or equal to -24,999,998.5 will make the original inequality \text{log}(-4x + 6) \geq 8 true.

Reflect and check

To check our solution, we can choose a number less than or equal to -24,999,998.5, for example, -25,000,000, and substitute it into the original inequality. We find that \text{log}(-4(-25,000,000) + 6) is indeed greater than or equal to 8, so our solution checks out. Remember, we have to ensure the argument of the logarithm (-4x + 6) is always positive, as the logarithm is only defined for positive numbers. If the check revealed that the argument of the logarithm was not positive, the solution would be considered extraneous and thus not a valid solution.

Idea summary

When solving logarithmic inequalities, we apply the properties of logarithms, convert the inequality to exponential form, and carefully consider the domain of the logarithm. Remember, the base of the logarithm can affect the direction of the inequality symbol. For bases greater than 1, the inequality symbol doesn't change direction, while for bases between 0 and 1, the inequality symbol flips. The domain of logarithmic functions only includes positive real numbers, which is important to keep in mind when identifying valid solutions.

Outcomes

2.13.A

Solve exponential and logarithmic equations and inequalities.

2.13.B

Construct the inverse function for exponential and logarithmic functions.

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