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מדריכי לימוד > College Algebra: Co-requisite Course

Logarithmic Equations*

Learning Objectives

  • Solve a logarithmic equation with algebra
  • Solve a logarithmic equation with a graph
  • Use the one-to-one property of logarithms to solve a logarithmic equation
  • Solve a radioactive decay problem
We have already seen that every logarithmic equation logb(x)=y{\mathrm{log}}_{b}\left(x\right)=y is equivalent to the exponential equation by=x{b}^{y}=x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log2(2)+log2(3x−5)=3{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x:

log2(2)+log2(3x−5)=3 log2(2(3x−5))=3Apply the product rule of logarithms. log2(6x−10)=3Distribute. 23=6x−10Apply the definition of a logarithm. 8=6x−10Calculate 23. 18=6xAdd 10 to both sides. x=3Divide by 6.\begin{array}{l}{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3\hfill & \hfill \\ \text{ }{\mathrm{log}}_{2}\left(2\left(3x - 5\right)\right)=3\hfill & \text{Apply the product rule of logarithms}.\hfill \\ \text{ }{\mathrm{log}}_{2}\left(6x - 10\right)=3\hfill & \text{Distribute}.\hfill \\ \text{ }{2}^{3}=6x - 10\hfill & \text{Apply the definition of a logarithm}.\hfill \\ \text{ }8=6x - 10\hfill & \text{Calculate }{2}^{3}.\hfill \\ \text{ }18=6x\hfill & \text{Add 10 to both sides}.\hfill \\ \text{ }x=3\hfill & \text{Divide by 6}.\hfill \end{array}

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression S and real numbers b and c, where b>0, b≠1b>0,\text{ }b\ne 1,

logb(S)=cif and only ifbc=S{\mathrm{log}}_{b}\left(S\right)=c\text{if and only if}{b}^{c}=S

Example: Using Algebra to Solve a Logarithmic Equation

Solve 2lnx+3=72\mathrm{ln}x+3=7.

Answer: 2lnx+3=7 2lnx=4Subtract 3. lnx=2Divide by 2. x=e2Rewrite in exponential form.\begin{array}{l}2\mathrm{ln}x+3=7\hfill & \hfill \\ \text{ }2\mathrm{ln}x=4\hfill & \text{Subtract 3}.\hfill \\ \text{ }\mathrm{ln}x=2\hfill & \text{Divide by 2}.\hfill \\ \text{ }x={e}^{2}\hfill & \text{Rewrite in exponential form}.\hfill \end{array}

Try It

Solve 6+lnx=106+\mathrm{ln}x=10.

Answer: x=e4x={e}^{4}

Example: Using a Graph to Understand the Solution to a Logarithmic Equation

Solve lnx=3\mathrm{ln}x=3.

Answer: lnx=3x=e3Use the definition of the natural logarithm.\begin{array}{l}\mathrm{ln}x=3\hfill & \hfill \\ x={e}^{3}\hfill & \text{Use the definition of the natural logarithm}\text{.}\hfill \end{array} Below is a graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words e3≈20{e}^{3}\approx 20. A calculator gives a better approximation: e3≈20.0855{e}^{3}\approx 20.0855.

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3). The graphs of y=lnxy=\mathrm{ln}x and y = 3 cross at the point (e3,3)\left(e^3,3\right), which is approximately (20.0855, 3).

Try It

Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2x=1000{2}^{x}=1000 to 2 decimal places.

Answer: x≈9.97x\approx 9.97

Use the one-to-one property of logarithms to solve logarithmic equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b≠1b\ne 1,

logbS=logbT if and only if S=T{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T.

For example,

If log2(x−1)=log2(8),then x−1=8\text{If }{\mathrm{log}}_{2}\left(x - 1\right)={\mathrm{log}}_{2}\left(8\right),\text{then }x - 1=8.

So, if x−1=8x - 1=8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log2(9−1)=log2(8)=3{\mathrm{log}}_{2}\left(9 - 1\right)={\mathrm{log}}_{2}\left(8\right)=3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation log(3x−2)−log(2)=log(x+4)\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x:

log(3x−2)−log(2)=log(x+4) log(3x−22)=log(x+4)Apply the quotient rule of logarithms. 3x−22=x+4Apply the one to one property of a logarithm. 3x−2=2x+8Multiply both sides of the equation by 2. x=10Subtract 2x and add 2.\begin{array}{l}\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)\hfill & \hfill \\ \text{ }\mathrm{log}\left(\frac{3x - 2}{2}\right)=\mathrm{log}\left(x+4\right)\hfill & \text{Apply the quotient rule of logarithms}.\hfill \\ \text{ }\frac{3x - 2}{2}=x+4\hfill & \text{Apply the one to one property of a logarithm}.\hfill \\ \text{ }3x - 2=2x+8\hfill & \text{Multiply both sides of the equation by }2.\hfill \\ \text{ }x=10\hfill & \text{Subtract 2}x\text{ and add 2}.\hfill \end{array}

To check the result, substitute x = 10 into log(3x−2)−log(2)=log(x+4)\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).

log(3(10)−2)−log(2)=log((10)+4) log(28)−log(2)=log(14) log(282)=log(14)The solution checks.\begin{array}{l}\mathrm{log}\left(3\left(10\right)-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(\left(10\right)+4\right)\hfill & \hfill \\ \text{ }\mathrm{log}\left(28\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(14\right)\hfill & \hfill \\ \text{ }\mathrm{log}\left(\frac{28}{2}\right)=\mathrm{log}\left(14\right)\hfill & \text{The solution checks}.\hfill \end{array}

A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

For any algebraic expressions S and T and any positive real number b, where b≠1b\ne 1,

logbS=logbT if and only if S=T{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How To: Given an equation containing logarithms, solve it using the one-to-one property.

  1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form logbS=logbT{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T.
  2. Use the one-to-one property to set the arguments equal.
  3. Solve the resulting equation, S = T, for the unknown.

Example: Solving an Equation Using the One-to-One Property of Logarithms

Solve ln(x2)=ln(2x+3)\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right).

Answer:  ln(x2)=ln(2x+3) x2=2x+3Use the one-to-one property of the logarithm. x2−2x−3=0Get zero on one side before factoring.(x−3)(x+1)=0Factor using FOIL. x−3=0 or x+1=0If a product is zero, one of the factors must be zero. x=3 or x=−1Solve for x.\begin{array}{l}\text{ }\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right)\hfill & \hfill \\ \text{ }{x}^{2}=2x+3\hfill & \text{Use the one-to-one property of the logarithm}.\hfill \\ \text{ }{x}^{2}-2x - 3=0\hfill & \text{Get zero on one side before factoring}.\hfill \\ \left(x - 3\right)\left(x+1\right)=0\hfill & \text{Factor using FOIL}.\hfill \\ \text{ }x - 3=0\text{ or }x+1=0\hfill & \text{If a product is zero, one of the factors must be zero}.\hfill \\ \text{ }x=3\text{ or }x=-1\hfill & \text{Solve for }x.\hfill \end{array}

Analysis of the Solution

There are two solutions: x = 3 or x = –1. The solution x = –1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Try It

Solve ln(x2)=ln1\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}1.

Answer: x=1x=1 or x=–1x=–1

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