Exponentials and logarithms: Year 4

Late September, and in Year 4 at vocational schools (istituti professionali) the maths syllabus is opening, just these days, the chapter on exponentials and logarithms, between the end of September and October. The first test arrives in five to ten weeks, so between the end of October and the end of November: it is not around the corner, but not so far off that you can forget about it now.

At vocational school this block is tackled with an applied slant: no proofs of the properties, lots of procedure and lots of checking the result. It is exactly what you need to pass the test, so it is what I show you here, with two exercises worked through to the final number.

What a logarithm is, in one sentence

The whole theory of logarithms rests on a single equivalence. If you have this clear, you solve half the exercises in the test:

lo g_{a} b = c ⟺ a^{c} = b

With $a > 0$ , $a \neq = 1$ , $b > 0$ . In words: the logarithm to base $a$ of $b$ is the exponent to which you must raise $a$ to get $b$ . $lo g_{2} 8 = 3$ because $2^{3} = 8$ , full stop. There is nothing else to memorise to understand what you are asking when you write a logarithm: you are asking “to what exponent?”.

From this definition also comes the condition you will use throughout the test: the argument of a logarithm must be positive. $lo g_{a} b$ makes sense only if $b > 0$ , whatever $a$ is. Keep it in mind, because you will apply it in Exercise 2.

It is also worth fixing two values that come back in almost every exercise, and that follow from the same definition without a calculator: $lo g_{a} 1 = 0$ for any base $a$ , because $a^{0} = 1$ ; and $lo g_{a} a = 1$ , because $a^{1} = a$ . If in a test you find a $lo g_{a} 1$ or a $lo g_{a} a$ “on its own” in the middle of a more complicated expression, that part simplifies at once, before applying any other property.

The three properties (and the mistake that costs the most marks)

The book presents three, and in a test those are enough:

lo g_{a} (x \cdot y) = lo g_{a} x + lo g_{a} y

lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y

lo g_{a} (x^{k}) = k \cdot lo g_{a} x

To these is added the change of base, useful when the calculator has only the “log” key (base 10) or “ln” (base $e$ ) and you need another base:

lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}

The mistake I see most often — and it is the point flagged as critical for this syllabus — is confusing the logarithm of a sum with the sum of two logarithms. There is no property for $lo g_{a} (x + y)$ . It doesn’t split, it doesn’t simplify, it stays as it is. A numerical counterexample settles the matter better than any rule learned by heart:

lo g (2 + 8) = lo g 10 = 1

lo g 2 + lo g 8 = 0.301 + 0.903 = 1.204

$1 \neq = 1.204$ . The properties hold for the product, quotient and power of the argument — not for a sum or difference inside the logarithm. If in an exercise you find yourself “distributing” a logarithm over a plus or a minus, you are wrong before you even do the calculations.

Exponential equations: when looking is enough, when you need the logarithm

When you can write both sides with the same base, the equation is solved without logarithms, by comparing only the exponents:

2^{x + 1} = 2^{5} ⟹ x + 1 = 5 ⟹ x = 4

The problem is when the bases can’t be made equal — because the right-hand side is not a clean power of the base on the left. In that case the only way to “bring down” the exponent is to apply the logarithm to both sides.

Exercise 1 — exponential equation solved with the logarithm

Solve:

5^{2 x - 1} = 40

$40$ is not an integer power of $5$ ( $5^{2} = 25$ , $5^{3} = 125$ : it falls in between), so comparing bases doesn’t work. I apply the natural logarithm to both sides:

ln (5^{2 x - 1}) = ln 40

By the power property, the exponent comes down in front of the logarithm:

(2 x - 1) \cdot ln 5 = ln 40

I isolate $2 x - 1$ :

2 x - 1 = \frac{ln 40}{ln 5} = \frac{3.6889}{1.6094} \approx 2.2920

2 x = 3.2920 ⟹ x \approx 1.646

Quick check: $5^{2 \cdot 1.646 - 1} = 5^{2.292} \approx 39.9$ , which rounds to $40$ . The result holds.

Before moving on to Exercise 2, if you want to practise on more equations of this kind you’ll find the Year 4 maths practice test on exponentials and logarithms here: maths test — exponentials and logarithms. It is calibrated on the high school (liceo) syllabus, so some questions are more technical than what you will be asked at vocational school: use it for the first ten questions, which are on the same basis as this article, and don’t worry if further on you find something you haven’t yet done in class.

Logarithmic equations: the condition of existence first of all

Here the typical mistake is not one of calculation: it is skipping the check on the argument. Every time you solve an equation with logarithms, before writing any algebraic step you must impose that every argument be positive. Only then can you apply the properties and solve. And at the end, you must check that the solutions found respect that condition — otherwise they must be discarded, even if algebraically they “come out right”.

Exercise 2 — logarithmic equation with a check of the condition of existence

Solve (logarithm to base 10):

lo g x + lo g (x - 3) = 1

Condition of existence. You need $x > 0$ and $x - 3 > 0$ , so:

x > 3

Solution. I use the product property to combine the two logarithms:

lo g [x (x - 3)] = 1

A base-10 logarithm equal to 1 means the argument is $1 0^{1}$ :

x (x - 3) = 10

x^{2} - 3 x - 10 = 0

I factorise: I look for two numbers that multiply to $- 10$ and add to $- 3$ : they are $- 5$ and $2$ .

(x - 5) (x + 2) = 0 ⟹ x = 5 or x = - 2

Check against the condition of existence. The condition was $x > 3$ . $x = 5$ respects it, $x = - 2$ doesn’t: it must be discarded. If you don’t check this step, you hand in an equation with two solutions when only one exists.

x = 5

Direct check: $lo g 5 + lo g 2 = lo g (5 \cdot 2) = lo g 10 = 1$ . Correct.

Two applications teachers often ask for in tests

Growth of a cost over time

The cost of a raw ingredient, in a restaurant, starts at $€ 25, 000$ a year and grows by $8%$ every year because of price increases. After how many years does it exceed $€ 40, 000$ ?

25, 000 \cdot 1.0 8^{t} = 40, 000

1.0 8^{t} = \frac{40 , 000}{25 , 000} = 1.6

I apply the base-10 logarithm to both sides:

t = \frac{lo g 1.6}{lo g 1.08} = \frac{0.2041}{0.0334} \approx 6.1 years

In a little over six years the cost exceeds $€ 40, 000$ , with the same annual growth.

Logarithmic scale: pH

In the kitchen and in food and wine studies, pH measures the acidity of a liquid with the formula $pH = - lo g_{10} [H^{+}]$ , where $[H^{+}]$ is the concentration of hydrogen ions in moles per litre. If a diluted vinegar has $[H^{+}] = 3.2 \times 1 0^{- 4} mol/L$ :

pH = - lo g (3.2 \times 1 0^{- 4}) = - (lo g 3.2 + lo g 1 0^{- 4}) = - (0.505 - 4) = 3.495

pH about $3.5$ : acidic, as you would expect from a vinegar. It is the same principle as decibels for sound: a logarithmic scale is used when the values involved vary by different orders of magnitude, and it is handy because it compresses them into small, readable numbers.

Note something common to the two exercises just seen: in both the logarithm is not the problem, it is the tool that brings down an awkward exponent or that turns a huge (or tiny) number into a readable figure. If in a problem from your field you are asked “after how long” or “what is it worth on a logarithmic scale”, the logarithm appears almost always as the answer to one of these two questions, not as an end in itself.

Summary of the two mistakes to avoid

There is no $lo g_{a} (x + y) = lo g_{a} x + lo g_{a} y$ . The properties hold for the product, quotient, power of the argument — never for a sum or difference.
Every logarithmic equation has a condition of existence on the argument, to be imposed before solving and checked afterwards, on every solution found.

If you already tackled equations with linear systems in Year 2, the mechanism of “I solve, then check that the solution is acceptable” will be familiar: you’ll find it applied to the same principles in the post on equations, inequalities and systems in Year 2 at vocational school, and even further back in the post on monomials and polynomials in Year 1. If in Year 4 you are also reviewing basic trigonometry, the formula sheet of Year 4 maths covers that part of the syllabus.

To practise more on exponentials and logarithms you can go back to the same practice test: maths test — exponentials and logarithms — the correct answer is free, the guided solution unlocks with your first lesson. Remember that it is calibrated on the high school, so the first questions are the most useful for your test.

You’ll find all the subjects we cover, maths included, at /materie/matematica. If you want a hand on this topic before the test, the first chat is free → contact.

Exponentials and logarithms in Year 4 of high school