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6. Exponential & Logarithmic Functions
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United States of AmericaFL
Grade 912

6.06 Exponential and logarithmic equations

Lesson
Practice
Lesson

Concept summary

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. That is to say, a log function of the form \log_b\left(x\right) has domain x>0. We may find that, for some log equations, a solution resulting from the process of solving is extraneous because it results in a negative argument.

We can solve logarithmic equations, which have non-zero expressions on both sides, graphically by setting the expressions on both sides equal to y, and then finding their point(s) of intersection.

We can also solve exponential equations, by taking the logarithm of both sides. In some cases we can make the process easier by identifying integer terms that may be able to be written as an power with the same base as another term in the equation. If we can do this, taking the logarithm of both sides will often leave us with a simple equation to solve.

Worked examples

Example 1

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

a

Graph y=7\log_{10}\left(x+3\right).

Approach

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 translated 3 units to the left, and then dilated vertically by a factor of 7.

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

Solution

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b

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

Approach

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

Solution

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

Approach

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

Solution

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

Reflection

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.

Example 2

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

a

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

Approach

We can rewrite the left hand side using the corollary of the product law 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.

To test for extraneous solutions we want to check if any solution falls outside the domain of the logarithms. We cannot take the logarithm of a negative value, so the domain for this equation is \left(0, \infty\right).

Solution

\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 law
\displaystyle \log_6\left(x^2+9x\right)\displaystyle =\displaystyle 2Distributive property
\displaystyle x^2+9x\displaystyle =\displaystyle 6^2\log_b x=n \iff x=b^n
\displaystyle x^2+9x\displaystyle =\displaystyle 36Evaluate the square
\displaystyle x^2+9x-36\displaystyle =\displaystyle 0Subtract 36
\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.

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

x=3 is a valid solution.

Reflection

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

\displaystyle \log_6\left(3\right)+\log\left(3+9\right)\displaystyle =\displaystyle \log_2\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.

b

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

Approach

We can rewrite the left hand side using the corollary of the quotient law 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 law\log_b\left(x\right)=\log_b\left(y\right) \iff x=y

To test for extraneous solutions we want to check if any solution falls outside the domain of the logarithms. We cannot take the logarithm of a negative value, so the domain for this equation is \left(2, \infty\right).

Solution

\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)Corollary of quotient law
\displaystyle \frac{x+4}{x-2}\displaystyle =\displaystyle xEquality law
\displaystyle x+4\displaystyle =\displaystyle x\left(x-2\right)Multiply by x-2
\displaystyle x+4\displaystyle =\displaystyle x^2-2xDistributive property
\displaystyle 0\displaystyle =\displaystyle x^2-3x-4Subtract x+4
\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.

We now need to check if either is extraneous. As -1 <0, it falls outside the domain of the function and is therefore extraneous.

x=4 is a valid solution.

Reflection

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

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

which is as required.

Example 3

Solve the following equation for x:

2^{1-2x}=1024

Approach

We can see that 1024 is a power of 2, so by first rewriting this as a power with a base of 2, we can simplify the process needed to solve this equation.

Solution

\displaystyle 2^{1-2x}\displaystyle =\displaystyle 1024State the equation
\displaystyle 2^{1-2x}\displaystyle =\displaystyle 2^{10}Rewrite 1024 as a power of 2
\displaystyle \log_2{\left(2^{1-2x}\right)}\displaystyle =\displaystyle \log_2{\left(2^{10}\right)}Take the \log_2 of both sides of the equation
\displaystyle 1-2x\displaystyle =\displaystyle 10\log_b\left(b^x\right)=x
\displaystyle -2x\displaystyle =\displaystyle 9Subtract 9 from both sides
\displaystyle x\displaystyle =\displaystyle -4.5Divide both sides by -2

Reflection

When we arrive at an equation such as 2^{1-2x}=2^{10} where both terms have the same base, we can simplify the process of taking logarithms by equating the terms in the exponent.

In this case, we have two equal terms with a base of 2, which means their respective exponents must be equal. That is, 1-2x=10

Outcomes

MA.912.AR.5.2

Solve one-variable equations involving logarithms or exponential expressions. Interpret solutions as viable in terms of the context and identify any extraneous solutions.

MA.912.AR.5.9

Solve and graph mathematical and real-world problems that are modeled with logarithmic functions. Interpret key features and determine constraints in terms of the context.

MA.912.NSO.1.6

Given a numerical logarithmic expression, evaluate and generate equivalent numerical expressions using the properties of logarithms or exponents.

MA.912.NSO.1.7

Given an algebraic logarithmic expression, generate an equivalent algebraic expression using the properties of logarithms or exponents.

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