Worked examples · Sheet 1

Equations with one logarithm on each side

Theory used: Equality of logarithms with the same base Domain conditions

Exercise 1

Solve the equation

lo g_{2} (x - 1) = lo g_{2} (7 - x) .

Domain conditions:

{x - 1 > 0 7 - x > 0 ⟺ {x > 1 x < 7 ⟺ 1 < x < 7

x - 1 = 7 - x ⟺ 2 x = 8 ⟺ x = 4

The value $x = 4$ satisfies the domain conditions.

Solution: $x = 4$ .

The arguments of the two logarithms contain the unknown, so before solving we require both of them to be positive; the two conditions must be satisfied together and form the system, which gives $1 < x < 7$ . The two sides are logarithms to the same base and the logarithmic function is injective, therefore the equation is equivalent to the equality of the arguments $x - 1 = 7 - x$ , from which $x = 4$ . Since $4$ belongs to the interval $(1, 7)$ , the domain conditions are satisfied and $x = 4$ is the solution.

Exercise 2

Solve the equation

lo g_{3} (2 x + 1) = lo g_{3} (x + 4) .

Domain conditions:

{2 x + 1 > 0 x + 4 > 0 ⟺ ⎩ ⎨ ⎧ x > - \frac{1}{2} x > - 4 ⟺ x > - \frac{1}{2}

2 x + 1 = x + 4 ⟺ x = 3

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

Solution: $x = 3$ .

The system of domain conditions requires both arguments to be positive; the more restrictive condition is $x > - \frac{1}{2}$ , so the system is equivalent to this single inequality. By the injectivity of the logarithmic function the equation is equivalent to the equality of the arguments, a linear equation with solution $x = 3$ . Since $3 > - \frac{1}{2}$ , the solution is acceptable.

Exercise 3

Solve the equation

lo g_{5} (x^{2} - 3) = lo g_{5} (2 x) .

Domain conditions:

{x^{2} - 3 > 0 2 x > 0 ⟺ {x < - 3 or x > 3 x > 0 ⟺ x > 3

x^{2} - 3 = 2 x ⟺ x^{2} - 2 x - 3 = 0 ⟺ x = 3 or x = - 1

The value $x = 3$ satisfies the domain conditions; the value $x = - 1$ does not satisfy them and is discarded.

Solution: $x = 3$ .

The first condition of the system is a quadratic inequality: the trinomial $x^{2} - 3$ is positive outside the roots $\pm 3$ . The intersection with $2 x > 0$ , that is $x > 0$ , leaves only $x > 3$ . The equality of the arguments gives the quadratic equation $x^{2} - 2 x - 3 = 0$ , with solutions $x = 3$ and $x = - 1$ . Since $3 > 3$ , the value $x = 3$ is acceptable; for $x = - 1$ the argument $2 x = - 2$ is negative, so the logarithm on the right-hand side does not exist and the solution is discarded.