10.2 Logarithms, Exponential Equations and Financial Mathematics

Current nbt.ac.za MAT Test Content groups surds, logarithms, and exponents together, including the solution of exponential equations and financial calculations such as compound interest, appreciation, and future value. On the calculator-free MAT paper the arithmetic is deliberately simple. The skill being tested is whether you can write the correct expression, not whether you can produce a long decimal.

Independent OpenExamPrep teaching uses original principals, rates, and times here. The 2015 CETAP MAT school booklet includes a compound-interest exemplar whose point is that the options are expressions such as P(1 + i)^n rather than a rand-and-cent evaluation. This section builds a parallel skill with different numbers.

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Surds and exponents you must move by hand

The laws you need are the same ones from FET algebra, but MAT will not cue which law. Products add exponents when the bases match: a^m × a^n = a^(m+n). A power of a power multiplies: (a^m)^n = a^(mn). A rational exponent a^(m/n) is the n-th root of a^m. A surd is simplified when the radicand has no perfect-square factor and no fraction remains under the root.

Worked surd simplification. Simplify √50 + √18 − √8.

√50 = √(25×2) = 5√2, √18 = √(9×2) = 3√2, and √8 = √(4×2) = 2√2. The sum is (5 + 3 − 2)√2 = 6√2.

Worked rationalising. Write 1 / (√7 − √3) with a rational denominator.

Multiply numerator and denominator by √7 + √3: (√7 + √3) / (7 − 3) = (√7 + √3)/4.

MAT options often include the unsimplified conjugate, the conjugate with the wrong sign, and the original denominator left in place. Simplify fully before you look.

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Exponential equations: same base, then equate

If a^u = a^v and a > 0, a ≠ 1, then u = v. The modelling step is to rewrite every factor as a power of one integer.

Worked exponential equation. Solve 4^(x+3) = 8^(x−1).

Write 4 = 2^2 and 8 = 2^3. Then 2^(2(x+3)) = 2^(3(x−1)), so 2x + 6 = 3x − 3, hence x = 9.

Check: 4^12 = (2^2)^12 = 2^24 and 8^8 = (2^3)^8 = 2^24. The check is fast because you never evaluate the huge integers; you compare exponents.

A second pattern is to take logs when the bases cannot be matched quickly: if 5^x = 12, then x = log 12 / log 5. MAT is more likely to keep the answer as that quotient, or to use a base that does match, than to ask for a decimal.

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Logarithm laws and domain

For a > 0, a ≠ 1, and positive arguments:

• log_a (xy) = log_a x + log_a y
• log_a (x/y) = log_a x − log_a y
• log_a (x^n) = n log_a x
• change of base: log_a b = log_c b / log_c a

The inverse statements a^(log_a x) = x and log_a (a^x) = x are how you unwind an exponential.

Worked log equation. Solve log_3 (x − 1) + log_3 (x + 3) = 2, with x > 1.

Combine: log_3 [(x − 1)(x + 3)] = 2, so (x − 1)(x + 3) = 9. Then x^2 + 2x − 3 = 9, x^2 + 2x − 12 = 0, (x + 4)(x − 3) = 0. The root x = −4 is outside the domain. The only solution is x = 3.

Discarding the negative trial root is part of the item. An option that keeps x = −4 is probing whether you remember that log arguments must be positive.

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Financial mathematics: they want the expression

Let P be the principal, r the nominal annual interest rate as a decimal, m the number of compounding periods in a year, and t the time in years. Then the period rate is i = r/m and the number of periods is n = mt. The future value after compound growth is

A = P(1 + i)^n = P(1 + r/m)^(mt).

That is the expression MAT wants. You will not be asked to evaluate (1.04)^14 by hand.

Worked parallel (original numbers). R4 500 is invested at a nominal 8% per year, compounded every six months, for 7 years. The period rate is 0.08/2 = 0.04. The number of half-year periods is 7 × 2 = 14. The investment is worth 4500(1.04)^14 rand.

Wrong expressions that will sit next to that answer:

• 4500(1.08)^7 treats the compounding as annual.
• 4500(1.04)^7 uses the correct period rate but counts years instead of periods.
• 4500(1.08)^14 doubles both the rate and the number of periods.

Build i and n separately, then write P(1 + i)^n. Do not invent a decimal.

Appreciation uses the same growth formula. A plot bought for R180 000 that appreciates at 6% per year compound is worth 180000(1.06)^t after t years. Reducing-balance depreciation replaces the plus with a minus: a machine bought for R80 000 that depreciates at 12.5% per year on the reducing balance has value 80000(7/8)^t after t years, because 12.5% = 1/8 and 1 − 1/8 = 7/8. Simple interest is the linear model A = P(1 + rt), which is not the same as compound growth. After 3 years at 8% simple, the factor is 1.24; after 3 years at 8% compound annually, the factor is (1.08)^3. Those two expressions are different options on purpose.

Future value in the MAT list includes the compound amount of a single deposit. If a stem ever describes equal deposits at the end of each period, the geometric series sum P[((1+i)^n − 1)/i] may appear, still as an expression. For the standard single-amount item, stay with P(1 + i)^n.

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A compact attack for the financial MCQ

1. Circle the compounding phrase: annually, half-yearly, quarterly, monthly.
2. Keep i = r/m and n = mt as two small working facts, written only on the surface permitted for your test mode.
3. Decide growth (1 + i) versus decay (1 − i) versus simple (1 + rt).
4. Write P times that factor. Stop. Match the option that is that expression.

The numbers are chosen so that r/m is a terminating decimal or a tidy fraction (8% half-yearly gives 0.04; 12% monthly gives 0.01; 12.5% decay gives 7/8). If your i looks ugly, you have probably divided by the wrong m.

Before you move on

• Same-base exponential equations equate exponents; mixed bases are rewritten first.
• Log sums become log products; check the domain before you accept a root.
• MAT finance asks for P(1 + r/m)^(mt), not a rand amount from a calculator.
• Simple interest, compound growth, and reducing-balance decay are three different expressions.