Equations with Repeated Powers | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Combine terms that contain the same power

Use index laws to expose the repeated part. For instance, $2^{x+1}=2(2^x)$ and $2^{x-1}=2^x/2$ .

Worked example

Solve 2ˣ⁺¹ + 6(2ˣ⁻¹) = 12.

\begin{gathered}2(2^x)+3(2^x)=12\\2^x=\frac{12}{5}\\x=\frac{\ln(12/5)}{\ln2}\end{gathered}

Combine before taking logs: there is no rule that splits the logarithm of the original sum.

Your turnPractice

Solve $3^{x+1}-3^{x-1}=8$ .

Show worked answer

(3-\tfrac13)3^x=8\Rightarrow3^x=3

Answer

x=1

Keep only positive exponential values

Since $e^{2x}=(e^x)^2$ , an equation containing $e^{2x}$ and $e^x$ may be quadratic in $u=e^x$ . This time $u>0$ : an exponential value cannot be zero or negative.

Worked example

Solve e²ˣ − 5eˣ + 6 = 0.

\begin{gathered}u=e^x>0\\u^2-5u+6=0\\(u-2)(u-3)=0\end{gathered}

Both 2 and 3 are positive. Undo the substitution for each.

Answer

x=\ln2\quad\text{or}\quad x=\ln3

A quadratic calculator is a quick way to find the $u$ candidates. If its roots are convenient, write the matching factors and check them. For awkward roots, use the quadratic formula to retain exact values before taking logs.

Worked example

Solve 2(3²ˣ) + 7(3ˣ) = 15.

\begin{gathered}u=3^x>0\\2u^2+7u-15=0\\(2u-3)(u+5)=0\end{gathered}

Keep $u=3/2$ ; reject $u=-5$ .

Answer

x=\frac{\ln(3/2)}{\ln3}\approx0.369

Your turnPractice

Solve $4^x+2^x-12=0$ . Why is $u=2^x$ a useful substitution?

Show worked answer

\begin{gathered}4^x=(2^x)^2\\u>0\\u^2+u-12=0\\(u+4)(u-3)=0\end{gathered}

Answer

x=\frac{\ln3}{\ln2}

Reject $u=-4$ ; an exponential is positive.

Clear a reciprocal power safely

$e^{-x}=1/e^x$ . Multiplying an equation by $e^x$ is safe because it is never zero.

Worked example

Solve eˣ − 6e⁻ˣ = 1.

\begin{gathered}e^{2x}-6=e^x\\u=e^x>0\\u^2-u-6=0\\(u-3)(u+2)=0\\x=\ln3\end{gathered}

Your turnPractice

Solve $e^x+2e^{-x}=3$ . Also explain why $2e^x+1=0$ has no real solution.

Show worked answer

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

Answer

x=0\quad\text{or}\quad x=\ln2

The second equation would require $e^x=-1/2$ , which is impossible.

Combine terms that contain the same power

Use index laws to expose the repeated part. For instance, $2^{x+1}=2(2^x)$ and $2^{x-1}=2^x/2$ .

Worked example

Solve 2ˣ⁺¹ + 6(2ˣ⁻¹) = 12.

\begin{gathered}2(2^x)+3(2^x)=12\\2^x=\frac{12}{5}\\x=\frac{\ln(12/5)}{\ln2}\end{gathered}

Combine before taking logs: there is no rule that splits the logarithm of the original sum.

Your turnPractice

Solve $3^{x+1}-3^{x-1}=8$ .

Show worked answer

(3-\tfrac13)3^x=8\Rightarrow3^x=3

Answer

x=1

Keep only positive exponential values

Since $e^{2x}=(e^x)^2$ , an equation containing $e^{2x}$ and $e^x$ may be quadratic in $u=e^x$ . This time $u>0$ : an exponential value cannot be zero or negative.

Worked example

Solve e²ˣ − 5eˣ + 6 = 0.

\begin{gathered}u=e^x>0\\u^2-5u+6=0\\(u-2)(u-3)=0\end{gathered}

Both 2 and 3 are positive. Undo the substitution for each.

Answer

x=\ln2\quad\text{or}\quad x=\ln3

Worked example

Solve 2(3²ˣ) + 7(3ˣ) = 15.

\begin{gathered}u=3^x>0\\2u^2+7u-15=0\\(2u-3)(u+5)=0\end{gathered}

Keep $u=3/2$ ; reject $u=-5$ .

Answer

x=\frac{\ln(3/2)}{\ln3}\approx0.369

Your turnPractice

Solve $4^x+2^x-12=0$ . Why is $u=2^x$ a useful substitution?

Show worked answer

\begin{gathered}4^x=(2^x)^2\\u>0\\u^2+u-12=0\\(u+4)(u-3)=0\end{gathered}

Answer

x=\frac{\ln3}{\ln2}

Reject $u=-4$ ; an exponential is positive.

Clear a reciprocal power safely

$e^{-x}=1/e^x$ . Multiplying an equation by $e^x$ is safe because it is never zero.

Worked example

Solve eˣ − 6e⁻ˣ = 1.

\begin{gathered}e^{2x}-6=e^x\\u=e^x>0\\u^2-u-6=0\\(u-3)(u+2)=0\\x=\ln3\end{gathered}

Your turnPractice

Solve $e^x+2e^{-x}=3$ . Also explain why $2e^x+1=0$ has no real solution.

Show worked answer

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

Answer

x=0\quad\text{or}\quad x=\ln2

The second equation would require $e^x=-1/2$ , which is impossible.