Worked examples · Sheet 3

Equations using the properties of logarithms and by substitution

Theory used: Properties of logarithms Substitution method Domain conditions Logarithm equal to a number

Exercise 1

Solve the equation

lo g_{2} x + lo g_{2} (x - 2) = 3.

Domain conditions:

{x > 0 x - 2 > 0 ⟺ x > 2

lo g_{2} (x (x - 2)) = 3 ⟺ x (x - 2) = 2^{3} = 8

x^{2} - 2 x - 8 = 0 ⟺ x = 4 or x = - 2

The value $x = 4$ satisfies the domain conditions; the value $x = - 2$ does not satisfy them and is discarded.

Solution: $x = 4$ .

The system of domain conditions requires $x > 0$ and $x > 2$ together, so it is equivalent to $x > 2$ . On positive arguments the sum of logarithms is the logarithm of the product, therefore the left-hand side becomes $lo g_{2} (x (x - 2))$ ; by the definition of logarithm the argument equals $2^{3} = 8$ . The quadratic equation $x^{2} - 2 x - 8 = 0$ has solutions $x = 4$ and $x = - 2$ . Since $4 > 2$ , the value $x = 4$ is acceptable; for $x = - 2$ both arguments of the given equation are negative, so the solution is discarded.

Exercise 2

Solve the equation

lo g_{3} (x + 6) - lo g_{3} x = 1.

Domain conditions:

{x + 6 > 0 x > 0 ⟺ x > 0

lo g_{3} \frac{x + 6}{x} = 1 ⟺ \frac{x + 6}{x} = 3^{1} = 3

x + 6 = 3 x ⟺ 6 = 2 x ⟺ x = 3

The value $x = 3$ satisfies the domain conditions.

Solution: $x = 3$ .

The domain conditions form the system with $x + 6 > 0$ and $x > 0$ ; the second is the more restrictive, so the system is equivalent to $x > 0$ . On positive arguments the difference of logarithms is the logarithm of the quotient, therefore the left-hand side becomes $lo g_{3} \frac{x + 6}{x}$ , and by the definition of logarithm the argument equals $3$ . Since $x > 0$ , multiplying both sides by $x$ is legitimate and gives the linear equation $x + 6 = 3 x$ , with solution $x = 3$ , which is positive and therefore acceptable. The check confirms the result, because $lo g_{3} 9 - lo g_{3} 3 = 2 - 1 = 1$ .

Exercise 3

Solve the equation

(lo g_{2} x)^{2} - lo g_{2} x - 6 = 0.

The domain condition is $x > 0$ . Setting $t = lo g_{2} x$ :

t^{2} - t - 6 = 0 ⟺ t = 3 or t = - 2

lo g_{2} x = 3 ⟺ x = 2^{3} = 8, lo g_{2} x = - 2 ⟺ x = 2^{- 2} = \frac{1}{4}

Both values are positive, so they satisfy the domain condition.

Solutions: $x = 8$ and $x = \frac{1}{4}$ .

The equation contains $(lo g_{2} x)^{2}$ and $lo g_{2} x$ , so the substitution $t = lo g_{2} x$ transforms it into a quadratic equation in $t$ . Unlike the exponential substitution, $t$ can take any real value, therefore both solutions $t = 3$ and $t = - 2$ are acceptable. By the definition of logarithm, from $lo g_{2} x = t$ it follows that $x = 2^{t}$ : we obtain $x = 8$ and $x = \frac{1}{4}$ , both positive, so the domain condition is satisfied in both cases.