Quadratic Logarithmic Equations | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

A repeated logarithm can be the unknown

If the same logarithm appears squared and unsquared, give its whole value a temporary name. This turns the equation into a quadratic.

Worked example

Solve (ln x)² − 3 ln x + 2 = 0.

The original domain is $x>0$ . Let $u=\ln x$ ; then $(\ln x)^2=u^2$ .

\begin{gathered}u^2-3u+2=0\\(u-1)(u-2)=0\\u=1\quad\text{or}\quad u=2\end{gathered}

These are values of $\ln x$ , not of $x$ . Undo the logarithm for each.

Answer

x=e\quad\text{or}\quad x=e^2

Both are positive. Their logs, 1 and 2, each make the original quadratic zero.

A negative value of $u=\ln x$ is allowed. It corresponds to a positive $x<1$ , so do not reject it just for its sign.

Your turnPractice

Solve $(\log_2 x)^2+\log_2 x-6=0$ .

Show worked answer

\begin{gathered}u=\log_2 x\\(u+3)(u-2)=0\end{gathered}

Answer

x=\frac18\quad\text{or}\quad x=4

$u=-3$ gives $x=2^{-3}$ ; $u=2$ gives $x=2^2$ .

Make the repeated expression match

Sometimes a log law must be used first. For $x>0$ , $\log_{10}(x^2)=2\log_{10}x$ .

Worked example

Solve (log₁₀ x)² − log₁₀(x²) = 3.

\begin{gathered}u=\log_{10}x\\u^2-2u-3=0\\(u-3)(u+1)=0\end{gathered}

Answer

x=1000\quad\text{or}\quad x=0.1

Your turnPractice

Solve $(\ln x)^2+\ln(x^2)=3$ . Give both answers exactly.

Show worked answer

\begin{gathered}u^2+2u-3=0\\(u+3)(u-1)=0\end{gathered}

Answer

x=e^{-3}\quad\text{or}\quad x=e

The first term requires $x>0$ , so the power-law rewrite is valid throughout the original domain.

A repeated logarithm can be the unknown

If the same logarithm appears squared and unsquared, give its whole value a temporary name. This turns the equation into a quadratic.

Worked example

Solve (ln x)² − 3 ln x + 2 = 0.

The original domain is $x>0$ . Let $u=\ln x$ ; then $(\ln x)^2=u^2$ .

\begin{gathered}u^2-3u+2=0\\(u-1)(u-2)=0\\u=1\quad\text{or}\quad u=2\end{gathered}

These are values of $\ln x$ , not of $x$ . Undo the logarithm for each.

Answer

x=e\quad\text{or}\quad x=e^2

Both are positive. Their logs, 1 and 2, each make the original quadratic zero.

A negative value of $u=\ln x$ is allowed. It corresponds to a positive $x<1$ , so do not reject it just for its sign.

Your turnPractice

Solve $(\log_2 x)^2+\log_2 x-6=0$ .

Show worked answer

\begin{gathered}u=\log_2 x\\(u+3)(u-2)=0\end{gathered}

Answer

x=\frac18\quad\text{or}\quad x=4

$u=-3$ gives $x=2^{-3}$ ; $u=2$ gives $x=2^2$ .

Make the repeated expression match

Sometimes a log law must be used first. For $x>0$ , $\log_{10}(x^2)=2\log_{10}x$ .

Worked example

Solve (log₁₀ x)² − log₁₀(x²) = 3.

\begin{gathered}u=\log_{10}x\\u^2-2u-3=0\\(u-3)(u+1)=0\end{gathered}

Answer

x=1000\quad\text{or}\quad x=0.1

Your turnPractice

Solve $(\ln x)^2+\ln(x^2)=3$ . Give both answers exactly.

Show worked answer

\begin{gathered}u^2+2u-3=0\\(u+3)(u-1)=0\end{gathered}

Answer

x=e^{-3}\quad\text{or}\quad x=e

The first term requires $x>0$ , so the power-law rewrite is valid throughout the original domain.