$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"power_simplified"}],"connections":[{"to":"power_simplified","from":"power","color":"sky","label":"evaluate the power"}]},"order":2,"title":"Remove the logarithm","description":"Equating the arguments to the base power gives the third method mark (M1). Evaluating $2^1$ is part of the same mark-bearing line, not a second mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2=\\frac{\\htmlData{anchor=fraction}{x^2}}{5}$"},{"latex":"$$\\htmlData{anchor=squared}{x^2}=10$"},{"latex":"$$x=\\pm\\sqrt{10}$"},{"latex":"$$\\boxed{x=\\sqrt{10}}$"}],"connections":[{"to":"squared","from":"fraction","color":"orange","label":"multiply by 5"}]},"order":4,"title":"Isolate and select x","description":"Multiply both sides by 5 to obtain the squared equation, then take square roots. Since the original logarithm requires x > 0, reject the negative root. The final boxed result earns the accuracy mark (A1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=\\pm\\sqrt{10}$"},{"text":"Reject x = −√10 because x > 0."},{"latex":"$$\\boxed{x=\\sqrt{10}}$"}]},"order":5,"title":"Select the valid root","description":"The original logarithm requires x > 0, so reject the negative root. Selecting and boxing the positive root earns the accuracy mark (A1)."}]}],"generatedAt":"2026-07-20T16:29:02.364Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":{"percent":52}},{"id":"feae588b-ccc6-49d9-910b-e2c20ede5492","revisionId":"216b316d-2636-4920-9e3d-582d49074d9d","questionNumber":"6","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":7,"sourceTitle":"Question Set 14: Exponentials and Logarithms","calculatorUse":null,"parts":[{"id":"16ec7159-47a8-4632-9f37-b901aee9c9b2","label":"0","content":"Giving your answers to 2 decimal places, solve the simultaneous equations\n\n$$\\begin{aligned} e^{2y} &= x + 1 \\\\ \\ln(x - 2) &= 2y - 1 \\end{aligned}$$","markscheme":"- Rearrange second equation: $x-2 = e^{2y-1} = \\frac{e^{2y}}{e}$ M1\n- Substitute $e^{2y} = x+1$: $x-2 = \\frac{x+1}{e}$ M1\n- Solve for $x$: $ex - 2e = x + 1 \\Rightarrow x(e-1) = 2e+1 \\Rightarrow x = \\frac{2e+1}{e-1}$ M1\n- $x \\approx 3.75$ A1\n- Find $y$: $e^{2y} = 3.75 + 1 = 4.75$ M1\n- $2y = \\ln(4.75) \\Rightarrow y = \\frac{\\ln(4.75)}{2}$ M1\n- $y \\approx 0.78$ A1","marks":7,"order":0,"markingRubric":{"totalMarks":7,"points":[{"id":"p1","marks":1,"description":"Converts the log equation to exponential form.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes one equation into the other to eliminate a variable.","code":"M"},{"id":"p3","marks":1,"description":"Rearranges to isolate $x$.","code":"M"},{"id":"p4","marks":1,"description":"Calculates $x = 3.75$.","code":"A"},{"id":"p5","marks":1,"description":"Substitutes $x$ back to find $y$.","code":"M"},{"id":"p6","marks":1,"description":"Uses logs to solve for $y$.","code":"M"},{"id":"p7","marks":1,"description":"Calculates $y = 0.78$.","code":"A"}],"finalAnswers":["x=3.75, y=0.78"],"generatedAt":"2026-05-12T02:14:06.626Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"substitution","label":"Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"\\ln(x-2)=2y-1"},{"latex":"x-2=e^{2y-1}"},{"latex":"x-2=\\frac{e^{2y}}{e}"}]},"order":1,"title":"Rewrite the logarithm","description":"Exponentiate the second equation, then use the index law for subtracting powers. Displaying the full rearrangement earns the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"x-2=\\frac{e^{2y}}{e}"},{"latex":"x-2=\\frac{x+1}{e}"}]},"order":2,"title":"Substitute for e^{2y}","description":"Use the first equation to replace the exponential term with x+1. This substitution earns the next M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"ex-2e=x+1"},{"latex":"x(e-1)=2e+1"},{"latex":"x=\\frac{2e+1}{e-1}"},

Exponentials and Logarithms

Question 1

Exponentials and Logarithms Questions

Exponentials and Logarithms

Question 1

Exponentials and Logarithms Questions