Logarithms, Sequences & Series

A sequence is an ordered list of numbers. A series adds its terms. A logarithm asks which power gives a number.

Translate logs to powers where useful, but enforce positive arguments first. For sums, look for cancellation or pairing.

The sequence starts 3, 6, 9, 12. If the rule adds 3 each time, what comes next? 1. Compare neighbours: each increases by 3. 2. Apply the stated rule to 12. 3. 12 + 3 = 15.

Each arrow adds the same difference of 3, so the next term is 15. The list is a sequence; adding the terms produces a series. A logarithm is a different operation: log₂ 8 = 3 because 2³ = 8. Always identify which operation a question asks for.

Solve log base 2 of (x-1) + log base 2 of (x-3)=3. Domain requires x>3. Combine to log base 2 of [(x-1)(x-3)]=3, so x^2-4x+3=8. Thus x^2-4x-5=0 and (x-5)(x+1)=0. Candidates are 5 and -1; only 5 lies in the domain. Check: log2(4)+log2(2)=2+1=3.

A few terms alone can fit several rules. Use the rule supplied by the problem. The product rule applies to logarithms of positive factors. Solving the polynomial alone would admit an invalid answer.

A few terms can fit many invented patterns. Use the rule stated in the question when one is given. A real logarithm needs a positive input and a positive base other than 1.

Translate logs to powers where useful, but enforce positive arguments first. For sums, look for cancellation or pairing.

Solve log base 2 of (x-1) + log base 2 of (x-3)=3. Domain requires x>3. Combine to log base 2 of [(x-1)(x-3)]=3, so x^2-4x+3=8. Thus x^2-4x-5=0 and (x-5)(x+1)=0. Candidates are 5 and -1; only 5 lies in the domain. Check: log2(4)+log2(2)=2+1=3.

The product rule applies to logarithms of positive factors. Solving the polynomial alone would admit an invalid answer.

log_b(x)=y means b^y=x, with b>0, b≠1 and x>0 for real logarithms. In particular log_b(1)=0 and log_b(b)=1. A logarithm's base cannot be negative or zero. Rules For positive x,y: log_b(xy)=log_b(x)+log_b(y), log_b(x/y)=log_b(x)−log_b(y), and log_b(x^m)=m log_b(x) for real m. Change of base gives log_b(x)=log_c(x)/log_c(b) for a valid base c. Thus log_b(x)=1/log_x(b) additionally requires x≠1. For real nonzero n, log_(b^n)(x^m)=(m/n)log_b(x). For positive a, a^m a^n=a^(m+n), a^m/a^n=a^(m−n), (a^m)^n=a^(mn), and a^(−n)=1/a^n. Restrictions are needed when extending these identities to negative bases and fractional powers. For real square roots, √(ab)=√a√b for a,b≥0, and √(a²)=|a| rather than always a.

log₂(8×4)=3+2=5. There is no sum rule log₂(8+4)=log₂(8)+log₂(4). To rationalise 1/(√3+√2), multiply by (√3−√2)/(√3−√2), obtaining √3−√2 because the denominator becomes 3−2=1.

A surd is an irrational root expression such as √2; not every expression containing a root is irrational. √9=3 and √2×√2=2 are rational. Define the permitted inputs before simplifying logs or roots.

An arithmetic progression (AP) has constant consecutive difference d; a geometric progression (GP) has a constant multiplier r. A sequence lists terms; a series sums them. Use the rule stated, because a finite list can fit many invented patterns. Key formulas For an AP with first term a, T_n=a+(n−1)d and S_n=n[2a+(n−1)d]/2. For a GP, T_n=ar^(n−1), and S_n=a(1−r^n)/(1−r) when r≠1; when r=1, S_n=na. The infinite GP sum is a/(1−r) when |r|<1. For two positive numbers x,y: AM=(x+y)/2, GM=√(xy), HM=2xy/(x+y); AM≥GM≥HM and AM×HM=GM². A harmonic progression has reciprocals in AP where those reciprocals exist.

For 5,8,11,…, the tenth term is 32 and the first-ten sum is 10(5+32)/2=185. For 3,6,12,…, the first-four sum is 3(1−2⁴)/(1−2)=45. Adding or multiplying every AP term by a constant preserves an AP; division requires a nonzero constant. Arithmetic–geometric sums For S_n=Σ from k=0 to n−1 of (a+kd)b r^k and r≠1: S_n=ab(1−r^n)/(1−r)+db[r−nr^n+(n−1)r^(n+1)]/(1−r)². When |r|<1 the infinite sum is ab/(1−r)+dbr/(1−r)². This is an arithmetic–geometric series, not a plain GP unless its arithmetic factor is constant.

A constant difference refers to subtraction, not the sum of neighbouring terms. Infinite-sum formulas need convergence; an r=1 denominator cannot be used.