Solving Logarithmic Equations & Inequalities (College Board AP® Precalculus): Revision Note

Logarithmic equations & inequalities

How do I solve a logarithmic equation?

• A logarithmic equation is an equation that involves one or more logarithmic expressions containing the variable

  • E.g. ${\log }_{3}x=5$ or $\ln (x+2)+\ln (x-1)=\ln 10$

• The key tool for solving logarithmic equations is the inverse relationship between logarithmic and exponential functions

  • ${\log }_{b}x=c\iff x=b^c$

    • this includes the case with base $e$:

      • $\ln x=c\iff x=e^c$

  • This allows you to convert a logarithmic expression into an exponential one

• Method 1: Converting directly to exponential form

  • If the equation has a single logarithmic expression equal to a constant

    • then you can convert using the definition of a logarithm

  • E.g. to solve ${\log }_{5}x=3$

    • Rewrite in exponential form: $x=5^3=125$

  • E.g. to solve $\ln (x+3)=2$

    • Rewrite: $x+3=e^2$

    • Solve: $x=e^2-3$

• Method 2: Using the one-to-one property

  • If two logarithmic expressions with the same base are equal

    • their arguments must be equal

  • I.e. if ${\log }_{b}A={\log }_{b}B$, then $A=B$

    • provided $A>0$ and $B>0$

  • E.g. to solve ${\log }_{10}(x^2-3)={\log }_{10}\left(2x\right)$

    • Set the arguments equal: $x^2-3=2x$

    • Rearrange: $x^2-2x-3=0$

    • Factor: $(x-3)(x+1)=0$

    • So $x=3$ or $x=-1$

  • After solving in this way, check for extraneous solutions (i.e. invalid solutions)

    • This is discussed below

• Method 3: Combining logarithms first, then solving

  • If the equation has multiple logarithmic terms

    • you can use the properties of logarithms to combine them into a single logarithm, then solve

  • The key properties used are:

    • Product property: ${\log }_b\left(xy\right)={\log }_{b}x+{\log }_{b}y$

    • Quotient property: ${\log }_b\left(\frac{x}{y}\right)={\log }_{b}x-{\log }_{b}y$

    • Power property: ${\log }_b\left(x^n\right)=n{\log }_{b}x$

  • E.g. to solve ${\log }_{10}(x-1)+{\log }_{10}(x+3)={\log }_{10}(x+9)$

    • Use the product property on the left-hand side

      • ${\log }_{10}\left(\right(x-1\left)\right(x+3\left)\right)={\log }_{10}(x+9)$

    • Apply the one-to-one property

      • $(x-1)(x+3)=x+9$

    • Expand and simplify

      • $x^2+2x-3=x+9$

      • giving $x^2+x-12=0$

    • Factor

      • $(x+4)(x-3)=0$

    • So $x=-4$ or $x=3$

  • After solving in this way, check for extraneous solutions (i.e. invalid solutions)

    • This is discussed below

Why do logarithmic equations produce extraneous solutions?

• Logarithmic equations frequently produce extraneous solutions

  • these are values that satisfy the algebraic equation

  • but are not valid in the original logarithmic equation

• This happens because logarithms are only defined for positive arguments

  • I.e. ${\log }_{b}x$ is only defined when $x>0$

  • So the domain of any logarithmic expression restricts which values of $x$ are valid as solutions

• You must always check your solutions against the domain of the original equation

  • Substitute each solution back into every logarithmic expression in the original equation

  • If any argument becomes zero or negative, that solution is extraneous and must be rejected

• E.g. the equation ${\log }_{10}(x-1)+{\log }_{10}(x+3)={\log }_{10}(x+9)$ was solved above to find $x=-4$ or $x=3$

  • For $x=3$

    • ${\log }_{10}\left(2\right)+{\log }_{10}\left(6\right)={\log }_{10}\left(12\right)$

    • all arguments (2, 6 and 12) are positive, so $x=3$ is valid

  • For $x=-4$

    • ${\log }_{10}(-5)+{\log }_{10}(-1)={\log }_{10}\left(5\right)$

    • ${\log }_{10}(-5)$ and ${\log }_{10}(-1)$ are both undefined

    • so $x=-4$ is extraneous

  • The only valid solution is $x=3$

• Extraneous solutions can also arise from contextual limitations in applied problems

  • E.g. if the variable represents a physical quantity like time or concentration, additional constraints on what values those quantities might take may apply

What about equations involving a coefficient in front of a logarithm?

• The power property of logarithms means that a coefficient in front of a logarithm can be interpreted as an exponent inside the logarithm

  • $n{\log }_{b}x={\log }_b\left(x^n\right)$

• Using this to solve a logarithmic equation can affect the number of solutions found

  • And some of those solutions might be extraneous

• E.g. consider the equation $2{\log }_{3}x=4$

• One way to solve it is to divide both sides by 2 first

  • This gives ${\log }_{3}x=2$

    • so $x=3^2=9$

• Another approach is to use the power property first

  • This gives ${\log }_3\left(x^2\right)=4$

    • so $x^2=3^4=81$

    • giving $x=9$ or $x=-9$

  • Note that $x=-9$ is a valid solution to ${\log }_3\left(x^2\right)=4$

    • because $(-9{)}^2=81>0$

• However, the two approaches give different solutions because

  • ${\log }_{3}x$ is only defined for $x>0$

  • while ${\log }_3\left(x^2\right)$ is defined for all $x\ne 0$

• To be valid a solution must satisfy the equation as given in the question

  • So in this case only $x=9$ is a valid solution

  • because the original equation contains ${\log }_{3}x$, not ${\log }_3\left(x^2\right)$

How do I solve logarithmic inequalities?

• Logarithmic inequalities are solved using similar techniques to logarithmic equations

• A key principle here is that a logarithmic function is either always increasing or always decreasing

  • This depends on the value of the base $b$

    • If $\mathit{b}\mathbf{>}\mathbf{1}$, the logarithmic function ${\log }_{b}x$ is increasing

    • If $\mathbf{0}\mathbf{<}\mathit{b}\mathbf{<}\mathbf{1}$, the logarithmic function ${\log }_{b}x$ is decreasing

• This means that if $x=p$ is the solution to ${\log }_{b}x=k$

  • then for everywhere on one side of $p$

    • ${\log }_{b}x>k$ is true

  • and for everywhere on the other side of $p$

    • ${\log }_{b}x<k$ is true

  • ${\log }_{b}x$ being increasing or decreasing will tell you which side is which

• E.g. to solve $\ln (x-1)>3$

  • Start by solving as an equation $\ln (x-1)=3$

    • rewrite in exponential form: $x-1=e^3$

  • $e>1$ so $\ln \left(x-1\right)={\log }_e\left(x-1\right)$ is an increasing function

    • That means $\ln \left(x-1\right)>3$ for $x-1>e^3$

  • Add 1 to both sides for the final solution

    • $x>1+e^3$

• You must also ensure that all arguments remain positive throughout the solution

  • The domain restriction (that the argument inside a logarithm must be greater than 0) acts as an additional constraint that must be combined with the inequality solution

  • In the above example

    • $x>1+e^3\Rightarrow x-1>e^3>0$

    • so the solution is valid