P6.5Solving exponential equations

• M1 for taking logarithms: \((x + 2)\ln 7 = \ln 10\) or \(x + 2 = \log_{7} 10\)
• A1 for awrt \(-0.817\)

Worked solution: \(x + 2 = \log_{7} 10 = \frac{\ln 10}{\ln 7} = 1.1832\ldots\), so \(x = -0.817\).

• M1 for \(2x - 3 = \ln 3\)
• A1 for \(x = \frac{\ln 3 + 3}{2}\) oe

Worked solution: \(2x - 3 = \ln 3 \Rightarrow x = \frac{\ln 3 + 3}{2}\) (\(\approx 2.049\)).

• M1 for \(4x + 3 = \mathrm{e}^{2}\)
• A1 for \(x = \frac{\mathrm{e}^{2} - 3}{4}\) oe

Worked solution: \(4x + 3 = \mathrm{e}^{2} \Rightarrow x = \frac{\mathrm{e}^{2} - 3}{4}\) (\(\approx 1.097\)).

• M1 for taking logarithms of both sides: \(x\ln 3 = (x - 1)\ln 5\)
• M1 for expanding and collecting the terms in \(x\)
• M1 for using the laws of logarithms to combine each side into a single logarithm
• A1 for \(P = \frac{1}{5}\) and \(Q = \frac{3}{5}\) (or equivalent, e.g. \(P = 5\), \(Q = \frac{5}{3}\))

Worked solution: \(x\ln 3 = (x - 1)\ln 5 \Rightarrow x(\ln 3 - \ln 5) = -\ln 5\)
\(\Rightarrow x = \frac{\ln \left(\frac{1}{5}\right)}{\ln \left(\frac{3}{5}\right)}\), using \(q\ln b - p\ln a = \ln\frac{b^{q}}{a^{p}}\).

• B1 for awrt \(3.15\)

Worked solution: \(x = \frac{\ln \left(\frac{1}{5}\right)}{\ln \left(\frac{3}{5}\right)} = 3.1506\ldots = 3.15\) (3 s.f.)

• M1 for forming a 3-term quadratic in \(u = 3^{x}\): \(3u^2 - 10u + 3 = 0\)
• A1 for \(u = 3\) and \(u = \frac{1}{3}\)
• M1 for solving \(3^{x} = u\) using logarithms (or powers) for a positive value of \(u\)
• A1 for \(x = -1, \ 1\)

Worked solution: Let \(u = 3^{x}\). Write \(3^{2x + 1} = 3u^2\): \(3u^2 - 10u + 3 = 0\), so \(u = 3 \text{ or } u = \frac{1}{3}\).
\(3^{x} = \frac{1}{3} \Rightarrow x = -1\); \(3^{x} = 3 \Rightarrow x = 1\).