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A recurrence can look almost geometric except for an added constant. A carefully chosen shift can remove that constant. The method is worth understanding because it explains the formula instead of guessing it from a short list of terms.

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Generate enough terms to check your work

Let a₁ = 1 and a_n+1 = 2a_n + 3 for integers n ≥ 1. The first five terms are 1, 5, 13, 29 and 61.

These terms are not a geometric sequence: the ratios are not constant. Repeated substitution will find a small-index term, but a transformation can reveal the general structure.

Choose a shift that absorbs the constant

Define b_n = a_n + 3. Then b_n+1 = a_n+1 + 3 = 2a_n + 6 = 2b_n.

The new sequence is geometric, with first term b₁ = 4 and common ratio 2. Hence b_n = 4 × 2^n-1, and a_n = 4 × 2^n-1 - 3.

Check the initial value and recurrence

At n = 1, the formula gives 4 - 3 = 1. At n = 5, it gives 64 - 3 = 61. Both agree with the direct calculation.

For a full structural check, substituting the formula into 2a_n + 3 gives 8 × 2^n-1 - 3, which is the formula for a_n+1. Checking only a few numerical terms would not establish that identity for every n.

Understand how the shift was found

If b_n = a_n + c, requiring b_n+1 = 2b_n gives 3 + c = 2c, so c = 3. The shift is chosen to make the algebra work.

Keep the starting index visible: a sequence beginning at n = 0 would use a different initial-value formula. This original exercise develops one algebra technique for AMC 12 preparation; it is not an official problem or a prediction of the paper.

Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.

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Official information and further reading: Source 1

Algebra can produce candidate values that do not belong to the domain of the original logarithms. In AMC 12 preparation, record the restrictions before transforming the equation, then apply them again at the end.

This is an original teaching example, not an official AMC 12 problem.

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Write the domain before the equation changes

Solve log₂(x) + log₂(x - 2) = 3 for real x. Each logarithm needs a positive argument: x > 0 and x - 2 > 0. Together these require x > 2.

The base is 2, which is positive and not equal to 1. These conditions make the real logarithms well defined.

Combine the logarithms on that domain

For positive arguments, the sum becomes log₂[x(x - 2)] = 3. Therefore x(x - 2) = 2³ = 8.

Rearrange to x² - 2x - 8 = 0, or (x - 4)(x + 2) = 0. The algebraic candidates are 4 and -2.

Return to the original restrictions

Only 4 satisfies x > 2. Substitution confirms log₂(4) + log₂(2) = 2 + 1 = 3. The real solution is x = 4.

At -2, both original arguments are negative. Their product is positive, so the combined expression alone can hide the original restriction. That is why a final product condition cannot replace the two starting argument conditions.

Try a related problem and check the same issue

Solve log₂(x) + log₂(x - 6) = 4. The domain is x > 6. Combining gives x(x - 6) = 16, whose candidates are 8 and -2. Only 8 is valid; its logs are 3 and 1.

Keep three lines in your notes: original domain, transformed equation and final filtered solution. Before using a logarithm rule, state why its arguments meet the rule's conditions. This small habit also helps with rational and radical equations, where algebraic transformations need their own restrictions.

Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 LOGS. Include your school year, country or competition region, experience and the question you want to discuss.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1