Unknown Bases and Nested Logarithms | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

When the base is unknown, check the base

If $x$ appears below “log”, it is the base, not the input. The restrictions are now $x>0$ and $x\ne1$ .

Worked example

Solve 4 logₓ 2 − logₓ 4 = 2.

\begin{gathered}\log_x16-\log_x4=2,\\\log_x4=2,\\x^2=4.\end{gathered}

The index equation gives $x=\pm2$ . Only 2 is a valid base. Check: $4\log_2 2-\log_2 4=4-2=2$ .

Answer

x=2

Your turnPractice

Solve $\log_x 9=2$ and $\log_x 4=-2$ . Keep only valid bases.

Show worked answer

\begin{gathered}x^2=9\Rightarrow x=3,\\x^{-2}=4\Rightarrow x=\tfrac12.\end{gathered}

A positive base below 1 is valid; the negative square roots are not.

Undo nested logarithms from the outside

$\log_2(\log_3 x)$ means first calculate $\log_3 x$ , then take the base-2 log of that result. To solve an equation, undo these operations in reverse order.

Worked example

Solve log₂(log₃ x) = −1.

The outer input needs $\log_3 x>0$ . Positive powers of 3 are greater than 1, so $x>1$ .

\begin{gathered}\log_3 x=2^{-1}=\frac12\\x=3^{1/2}=\sqrt3\end{gathered}

It passes the domain check. The inner log is $1/2$ , whose base-2 log is −1.

Your turnPractice

Solve $\ln(\ln x)=0$ , checking both log inputs.

Show worked answer

Answer

x=e

$\ln x=e^0=1$ , so $x=e^1$ . Both inputs, $x=e$ and $\ln x=1$ , are positive.

When the base is unknown, check the base

If $x$ appears below “log”, it is the base, not the input. The restrictions are now $x>0$ and $x\ne1$ .

Worked example

Solve 4 logₓ 2 − logₓ 4 = 2.

\begin{gathered}\log_x16-\log_x4=2,\\\log_x4=2,\\x^2=4.\end{gathered}

The index equation gives $x=\pm2$ . Only 2 is a valid base. Check: $4\log_2 2-\log_2 4=4-2=2$ .

Answer

x=2

Your turnPractice

Solve $\log_x 9=2$ and $\log_x 4=-2$ . Keep only valid bases.

Show worked answer

\begin{gathered}x^2=9\Rightarrow x=3,\\x^{-2}=4\Rightarrow x=\tfrac12.\end{gathered}

A positive base below 1 is valid; the negative square roots are not.

Undo nested logarithms from the outside

$\log_2(\log_3 x)$ means first calculate $\log_3 x$ , then take the base-2 log of that result. To solve an equation, undo these operations in reverse order.

Worked example

Solve log₂(log₃ x) = −1.

The outer input needs $\log_3 x>0$ . Positive powers of 3 are greater than 1, so $x>1$ .

\begin{gathered}\log_3 x=2^{-1}=\frac12\\x=3^{1/2}=\sqrt3\end{gathered}

It passes the domain check. The inner log is $1/2$ , whose base-2 log is −1.

Your turnPractice

Solve $\ln(\ln x)=0$ , checking both log inputs.

Show worked answer

Answer

x=e

$\ln x=e^0=1$ , so $x=e^1$ . Both inputs, $x=e$ and $\ln x=1$ , are positive.