Exponential Inequalities | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Taking ln preserves order; dividing may not

The function $\ln x$ increases: larger positive inputs have larger logs. Taking ln of two positive sides keeps an inequality’s direction. Dividing by a negative number reverses it.

Worked example

Solve 0.6ˣ < 0.7 exactly.

\begin{gathered}x\ln0.6<\ln0.7\\\ln0.6<0\\x>\frac{\ln0.7}{\ln0.6}\end{gathered}

The reversal happens when dividing by $\ln0.6$ , not when taking ln. The boundary is approximately 0.698, but use the exact log ratio for the exact solution set.

Below the horizontal line means to the right of the boundary for this decreasing curve.

Check one value on each side: $x=1$ works since $0.6<0.7$ ; $x=0$ fails since $1>0.7$ . Equality is excluded because the question uses a strict inequality.

Your turnPractice

Solve $4\times3^{2x-1}\ge5$ exactly.

Show worked answer

\begin{gathered}3^{2x-1}\ge\tfrac54,\\(2x-1)\ln3\ge\ln(5/4),\\x\ge\frac{\ln(15/4)}{2\ln3}.\end{gathered}

Dividing by positive $2\ln3$ keeps the direction; the equality boundary is included.

Logarithmic inequalities still need a domain

Worked example

Solve ln(2x + 1) ≤ ln(x + 4).

Both arguments must be positive, giving $x>-1/2$ . As ln is increasing, compare its inputs.

2x+1\le x+4\Rightarrow x\le3

Both requirements must hold:

Answer

-\frac12<x\le3

Logs to bases between 0 and 1 are decreasing, so taking those logs reverses order. Using ln or base-10 logs avoids that extra reversal; you still check the sign when dividing.

For a quadratic inequality, factor first. A negative product needs one positive factor and one negative factor. For example, $(u-1)(u-4)<0$ holds between 1 and 4. If $u=e^x$ , keep $u>0$ , then take ln of each positive boundary to find the interval for $x$ .

Your turnPractice

Solve $e^{2x}-5e^x+6<0$ . Use $u=e^x$ and check an interior value.

Show worked answer

\begin{gathered}(u-2)(u-3)<0\\2<u<3\end{gathered}

Between the roots one factor is positive and the other negative. Outside them the product is positive.

Answer

\ln2<x<\ln3

At $x=\ln(5/2)$ , $u=2.5$ and the product is $(0.5)(-0.5)<0$ .

Taking ln preserves order; dividing may not

The function $\ln x$ increases: larger positive inputs have larger logs. Taking ln of two positive sides keeps an inequality’s direction. Dividing by a negative number reverses it.

Worked example

Solve 0.6ˣ < 0.7 exactly.

\begin{gathered}x\ln0.6<\ln0.7\\\ln0.6<0\\x>\frac{\ln0.7}{\ln0.6}\end{gathered}

The reversal happens when dividing by $\ln0.6$ , not when taking ln. The boundary is approximately 0.698, but use the exact log ratio for the exact solution set.

Below the horizontal line means to the right of the boundary for this decreasing curve.

Check one value on each side: $x=1$ works since $0.6<0.7$ ; $x=0$ fails since $1>0.7$ . Equality is excluded because the question uses a strict inequality.

Your turnPractice

Solve $4\times3^{2x-1}\ge5$ exactly.

Show worked answer

\begin{gathered}3^{2x-1}\ge\tfrac54,\\(2x-1)\ln3\ge\ln(5/4),\\x\ge\frac{\ln(15/4)}{2\ln3}.\end{gathered}

Dividing by positive $2\ln3$ keeps the direction; the equality boundary is included.

Logarithmic inequalities still need a domain

Worked example

Solve ln(2x + 1) ≤ ln(x + 4).

Both arguments must be positive, giving $x>-1/2$ . As ln is increasing, compare its inputs.

2x+1\le x+4\Rightarrow x\le3

Both requirements must hold:

Answer

-\frac12<x\le3

Logs to bases between 0 and 1 are decreasing, so taking those logs reverses order. Using ln or base-10 logs avoids that extra reversal; you still check the sign when dividing.

Your turnPractice

Solve $e^{2x}-5e^x+6<0$ . Use $u=e^x$ and check an interior value.

Show worked answer

\begin{gathered}(u-2)(u-3)<0\\2<u<3\end{gathered}

Between the roots one factor is positive and the other negative. Outside them the product is positive.

Answer

\ln2<x<\ln3

At $x=\ln(5/2)$ , $u=2.5$ and the product is $(0.5)(-0.5)<0$ .