Why study logarithmsMany public-sector entrance exams (in Brazil, concursos) do not allow calculators, and that is where logarithms come in: with them, we can compute compound interest amounts and terms using only a few values that the exam itself provides. Logarithms arose mainly as a tool for solving exponential equations, and today they appear in mathematical models in many fields.Their great usefulness is turning powers into multiplications and multiplications into sums. An exponential equation, where the unknown sits in the exponent, becomes a first-degree equation that can be solved with pencil and paper.Definition: given positive real numbers a and b, with a different from 1, the logarithm of b to base a is the exponent x to which a must be raised so that the resulting power equals b.\[log_a b = x <=> a^{x} = b\]In other words, a logarithm is nothing more than an exponent. We say that "a" is the base, "b" is the argument and "x" is the logarithm. When no base is written, it is 10 (the common logarithm, the one most used in exams); ln denotes the natural logarithm, whose base is e ≈ 2.718.Properties of logarithmsI) The logarithm of 1, in any base, equals zero;II) A logarithm whose base equals its argument equals one;III) In the same base, two logarithms are equal if and only if their arguments are equal;IV) Logarithm of a product:\[log_c(a · b) = log_c(a) + log_c(b)\]V) Logarithm of a quotient:\[log_c(\frac{a}{b}) = log_c(a) - log_c(b)\]VI) Logarithm of a power:\[log_c(a^{b}) = b · log_c(a)\]VII) Logarithm of a root:\[log_c(a^{\frac{1}{b}}) = \frac{log_c(a)}{b}\]With these properties, a few values given in the exam are enough for many calculations. For example, with log 2 = 0.30 and log 3 = 0.48, we get log 1.08 = log(2² · 3³) - log 100 = 0.60 + 1.44 - 2 = 0.04.Given log 2 = 0.30 and log 3 = 0.48, what is log 12?Answer: 1.08 — 12 = 2² · 3, so log 12 = 2 · log 2 + log 3 = 0.60 + 0.48 = 1.08. The option 0.78 comes from adding log 2 and log 3 while forgetting that the 2 is squared; 0.144 comes from multiplying the logarithms, which no property allows.Logarithms in financial mathematics and examsWe start from the familiar compound interest formula and take the logarithm of both sides:\[ M_t = C · (1 + i)^{t} \]\[ log(M_t) = log(C · (1 + i)^{t}) \]\[ log(M_t) = log(C) + log((1 + i)^{t}) \]\[ log(M_t) = log(C) + t · log(1 + i) \]The equation, which was exponential, becomes linear in t. In other words: each period that passes, the logarithm of the amount increases by the same quantity, log(1 + i), no matter what moment we are at.ExamplesAn amount of R$ 10,000.00 (R$ is the Brazilian real) was invested at compound interest of 2% per month for 125 months. Given that log(1.02) = 0.008, what is the final amount of this investment?\[ log(M_t) = log(C) + t · log(1 + i) \]\[ log(M_t) = log(10000) + 125 · log(1 + 0.02) \]\[ log(M_t) = 4 + 125 · 0.008 = 5 \]\[ M_t = 10^{5} = 100000 \]The final amount is R$ 100,000.00 with the given data. Note: 0.008 is a rough approximation (the true value of log 1.02 is 0.0086), chosen to make the arithmetic come out round. With the exact value, the amount would be about R$ 118,856.00. In the exam, always use the values the question provides.Júlio has an amount Q, in reais, and intends to invest it under compound interest at a rate of 2% per month. Given log 2 = 0.3010 and log 1.02 = 0.0086, how long will it take for this amount to quadruple?\[ log(M_t) = log(C) + t · log(1 + i) \]\[ log(4 · Q) = log(Q) + t · log(1 + 0.02) \]\[ log(4 · Q) - log(Q) = t · log(1.02) \]\[ log(\frac{4 · Q}{Q}) = t · log(1.02) \]\[ log(4) = t · log(1.02) \]\[ log(2^{2}) = t · 0.0086 \]\[ 2 · log(2) = t · 0.0086 \]\[ 2 · 0.3010 = t · 0.0086 \]\[ t = \frac{0.6020}{0.0086} = 70 \text{ months} = \text{5 years and 10 months} \]Notice that the value of Q does not matter: it cancels out in the division. The time needed to quadruple depends only on the rate.A sum is invested at compound interest of 2% per month. Using log 2 = 0.30 and log 1.02 = 0.0086, approximately how long does it take to double?Answer: 35 months — doubling means M/C = 2, so t = log 2 / log 1.02 = 0.30 / 0.0086 ≈ 34.9, that is, about 35 months. The option 70 months is the time to quadruple (doubling twice); 50 months is what simple interest would give.Calculating the term in daysWhen the rate is given for a period of m days (for example, 30 days) and we want the term n in days, the exponent becomes n/m. Let us find the term of a transaction knowing only the rate, the final amount and the principal:Interest rate: 1% every 30 days;Final amount: R$ 10,803.46;Principal: R$ 10,000.00.\[ M = C · (1 + i)^{\frac{n}{m}}\]\[ \frac{M}{C} = (1 + i)^{\frac{n}{m}}\]\[ ln(\frac{M}{C}) = \frac{n}{m} · ln(1 + i)\]\[ n = m · \frac{ln(\frac{M}{C})}{ln(1 + i)}\]Substituting the values:\[ n = 30 · \frac{ln(\frac{10803.46}{10000})}{ln(1 + 0.01)} = 233.0014\]The term between the loan date and the payment date is therefore 233 days. The formula works with either the natural logarithm (ln) or the common logarithm (log), as long as the same one is used in both parts of the fraction.A principal of R$ 10,000.00 invested at 1% every 30 days grew to R$ 10,201.00. What was the term of the investment?Answer: 60 days — M/C = 1.0201 = 1.01², so n/30 = 2 and n = 60 days. No logarithm is even needed: just recognize 1.0201 as 1.01 squared. The 0.01% above 2% is the interest on interest of the second period, not a third period.Exercises1) An investment yields compound interest at a rate of 6% per year. After how many years will an initial amount of R$ 1,000.00 reach R$ 10,000.00? (Use log(1.06) = 0.025.)2) An investor bought a block of a company's shares for R$ 1,000.00 and sold it n months later for R$ 4,000.00. Assuming the shares appreciated 8% per month and using the approximations log 2 = 0.3 and log 3 = 0.48, what is the value of n?3) Saulo invested R$ 45,000.00 in an investment fund that yields 20% per year. His goal is to use the proceeds to buy a house that, on the investment date, cost R$ 135,000.00 and appreciates at 8% per year. Under these conditions, how many years after the investment date will Saulo be able to buy the house? (Use the approximation log 3 = 0.48.)4) (PUC-SP) A principal C, invested at compound interest at a rate i per period, produces at the end of n periods the amount M given by M = C · (1 + i)^n. Under these conditions, using log 2 = 0.30 and log 3 = 0.48, after how long will a principal of R$ 2,000.00, invested at compound interest at 20% per year, grow to R$ 5,000.00?5) (CESGRANRIO-2008) An investment of R$ 1,000.00 was made at compound interest of 3% per month. After a period of t months, the amount was R$ 1,159.27. What is the value of t? (Data: ln(1,000) = 6.91; ln(1,159.27) = 7.06; ln(1.03) = 0.03.)6) (ESPM) An amount of R$ 10,000.00 was invested at compound interest of 4% per month for 10 months. Given that log 1.04 = 0.017 and log 1.48 = 0.17, how much interest did this investment earn?7) (Calculator allowed) To pay off a loan of R$ 175,000.00, R$ 197,338.77 was paid. With an interest rate of 2% every 30 days, find the term of this loan.8) (Calculator allowed) An account pays the return of a stock portfolio whose historical return is 4.20% every 30 days. How long will it take for each R$ 1.00 deposited to become R$ 2.12023?9) (Calculator allowed) Find how long it takes for a principal to earn 40% interest if invested at a rate of 35.02% every 365 days.Answer key1) 40 years.2) 15 months.3) 12 years.4) 5 years.5) t = 5 months.6) R$ 4,800.00.7) 182 days.8) 548 days.9) 409 days.
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