Solving Logarithmic Equations | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Isolate one logarithm, then undo it

The argument of a logarithm is its whole input. Before changing an equation, require every original argument to be positive.

Worked example

Solve ln(2x − 1) = 3.

The domain is $2x-1>0$ , so $x>1/2$ . The log asks for an exponent of $e$ ; therefore

\begin{gathered}2x-1=e^3\\x=\frac{e^3+1}{2}\end{gathered}

This means raise $e$ to both sides, not multiply both sides by $e$ . The result is greater than $1/2$ , as required.

Worked example

Solve ln(x + 2) − ln x = 1.

The original arguments give $x>0$ . Combine first, then undo the log.

\begin{gathered}\ln\left(\frac{x+2}{x}\right)=1,\\x+2=ex,\\x=\frac2{e-1}.\end{gathered}

The answer is positive and makes the ratio $e$ , whose natural log is 1.

Your turnPractice

Solve $\ln(3x-2)=2$ exactly, and check the original argument.

Show worked answer

Answer

x=\frac{e^2+2}{3}

It makes $3x-2=e^2>0$ , satisfying the original domain $x>2/3$ .

Equal logs have equal positive inputs

For the same valid base, each positive input has one distinct logarithm. Thus $\log_a U=\log_a V$ means $U=V$ , provided both inputs are positive.

Worked example

Solve 2 log₈(x + 2) = log₈(2x + 19).

The original domain is $x>-2$ : this makes both arguments positive. Apply the power law, then equate the inputs.

\begin{gathered}(x+2)^2=2x+19,\\x^2+2x-15=0,\\(x-3)(x+5)=0.\end{gathered}

The candidates are 3 and −5. A candidate is an algebraic result still needing the original checks. Reject −5: it makes $x+2=-3$ . At 3, both sides equal $\log_8 25$ .

Answer

x=3

A root of the transformed equation is not automatically a root of the original log equation.

Your turnPractice

Solve $\ln(x-1)+\ln(x+1)=\ln8$ . Why is one algebraic root invalid?

Show worked answer

\begin{gathered}x>1\\(x-1)(x+1)=8\\x^2=9\\x=\pm3\end{gathered}

Answer

x=3

At −3 the original log inputs are negative, even though their product is positive. At 3, the product is $2\times4=8$ .

Isolate one logarithm, then undo it

The argument of a logarithm is its whole input. Before changing an equation, require every original argument to be positive.

Worked example

Solve ln(2x − 1) = 3.

The domain is $2x-1>0$ , so $x>1/2$ . The log asks for an exponent of $e$ ; therefore

\begin{gathered}2x-1=e^3\\x=\frac{e^3+1}{2}\end{gathered}

This means raise $e$ to both sides, not multiply both sides by $e$ . The result is greater than $1/2$ , as required.

Worked example

Solve ln(x + 2) − ln x = 1.

The original arguments give $x>0$ . Combine first, then undo the log.

\begin{gathered}\ln\left(\frac{x+2}{x}\right)=1,\\x+2=ex,\\x=\frac2{e-1}.\end{gathered}

The answer is positive and makes the ratio $e$ , whose natural log is 1.

Your turnPractice

Solve $\ln(3x-2)=2$ exactly, and check the original argument.

Show worked answer

Answer

x=\frac{e^2+2}{3}

It makes $3x-2=e^2>0$ , satisfying the original domain $x>2/3$ .

Equal logs have equal positive inputs

For the same valid base, each positive input has one distinct logarithm. Thus $\log_a U=\log_a V$ means $U=V$ , provided both inputs are positive.

Worked example

Solve 2 log₈(x + 2) = log₈(2x + 19).

The original domain is $x>-2$ : this makes both arguments positive. Apply the power law, then equate the inputs.

\begin{gathered}(x+2)^2=2x+19,\\x^2+2x-15=0,\\(x-3)(x+5)=0.\end{gathered}

The candidates are 3 and −5. A candidate is an algebraic result still needing the original checks. Reject −5: it makes $x+2=-3$ . At 3, both sides equal $\log_8 25$ .

Answer

x=3

A root of the transformed equation is not automatically a root of the original log equation.

Your turnPractice

Solve $\ln(x-1)+\ln(x+1)=\ln8$ . Why is one algebraic root invalid?

Show worked answer

\begin{gathered}x>1\\(x-1)(x+1)=8\\x^2=9\\x=\pm3\end{gathered}

Answer

x=3

At −3 the original log inputs are negative, even though their product is positive. At 3, the product is $2\times4=8$ .