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Solving Logarithmic Equations

A logarithmic equation is an equation containing one or more logarithms containing a variable in the argument of the logarithmic function.

It can be expressed as:
logbx=k\boxed{{\log_{b}{x}=k}}
where kRk\in\mathbb{R}.

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How to Solve Logarithmic Equations

  1. Combine like terms and express as a single logarithm.
  2. Exponentiate both sides of the equation.
  3. Solve for x.
  4. Determine if there are any extraneous solutions by verifying the answers.

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Example

log2(x1)=2\log_{2}(x-1)=2


Step 1.
Already completed.


Step 2/3.

log2(x1)=22log2(x1)=22x1=4x=5\begin{array}{rcl} \log_{2}(x-1)&=&2\\\\ 2^{\log_{2}(x-1)}&=&2^2\\\\ x-1&=&4\\\\ x&=&5 \end{array}


Step 4.
Verify.

log2(x1)=2log2(51)=2log24=22=2\begin{array}{rcl} \log_{2}(x-1)&=&2\\\\ \log_{2}(5-1)&=&2\\\\ \log_{2}4&=&2\\\\ 2&=&2&&\color{red}\checkmark \end{array}


Therefore, x=5x=5 is the solution.
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Example: Solving Logarithmic Equations

Solve and check.

log(x+2)+log(x1)=1\log(x+2)+\log(x-1)=1


log(x+2)+log(x1)=1log(x+2)(x1)=1(x2+x2)=101x2+x210=0x2+x12=0(x+4)(x3)=0 x=4,3\begin{array}{rcl} \log(x+2)+\log(x-1)&=&1\\\\ \log(x+2)(x-1)&=&1\\\\ (x^2+x-2)&=&10^1\\\\ x^2+x-2-10&=&0\\\\ x^2+x-12&=&0\\\\ (x+4)(x-3)&=&0\\\\ \therefore~x&=&-4, 3 \end{array}

Check.

log(x+2)+log(x1)=1log(3+2)+log(31)=1log(5)+log(2)=1log10=11=1\begin{array}{rcl} \log(x+2)+\log(x-1)&=&1\\\\ \log(3+2)+\log(3-1)&=&1\\\\ \log(5)+\log(2)&=&1\\\\ \log10&=&1\\\\ 1&=&1&&\color{red}\checkmark \end{array} log(x+2)+log(x1)=1log(4+2)+log(41)=1log(2)+log(5)=1 x=4  is an extraneous solution\begin{array}{rcl} \log(x+2)+\log(x-1)&=&1\\\\ \log(-4+2)+\log(-4-1)&=&1\\\\ \log(-2)+\log(-5)&=&1\\\\ \therefore~x&=&-4~~\text{is an extraneous solution} \end{array}


Therefore, x=3x=3 is the only solution.

Practice: Solving Logarithmic Equations

Solve and check.

log5(x+1)+log5(x3)=1\log_{5}{(x+1)}+\log_{5}{(x-3)}=1

Practice: Solving Logarithmic Equations

Solve.

log2x+log4x=3\log_{2}{x}+\log_{4}{x}=3

Practice: Solving Logarithmic Equations

If log41024=a+b\log_{4}{1024}=a+b and log864=ab\log_{8}{64}=a-b, what is aa and bb?
Extra Practice