Worked examples · Sheet 2

Equations with the logarithm equal to a number

Theory used: Logarithm equal to a number Logarithm

Exercise 1

Solve the equation

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

lo g_{2} (x + 3) = 4 ⟺ x + 3 = 2^{4} = 16 ⟺ x = 13

The domain condition $x + 3 > 0$ is satisfied, since $x + 3 = 16 > 0$ .

Solution: $x = 13$ .

The logarithm to base $2$ of $x + 3$ equals $4$ , so by the definition of logarithm the argument is the power $2^{4} = 16$ . What remains is the linear equation $x + 3 = 16$ , with solution $x = 13$ . The domain condition is automatically satisfied, because the argument turns out to be equal to $16$ , which is positive.

Exercise 2

Solve the equation

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

lo g_{3} (2 x - 1) = 2 ⟺ 2 x - 1 = 3^{2} = 9 ⟺ x = 5

The domain condition $2 x - 1 > 0$ is satisfied, since $2 x - 1 = 9 > 0$ .

Solution: $x = 5$ .

By the definition of logarithm, the argument $2 x - 1$ is equal to the base $3$ raised to the value of the logarithm, that is $3^{2} = 9$ . The linear equation $2 x - 1 = 9$ gives $2 x = 10$ , so $x = 5$ . The check confirms the result, because $lo g_{3} (2 \cdot 5 - 1) = lo g_{3} 9 = 2$ .

Exercise 3

Solve the equation

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

lo g_{1/2} (x + 1) = - 3 ⟺ x + 1 = (\frac{1}{2})^{- 3} = 2^{3} = 8 ⟺ x = 7

The domain condition $x + 1 > 0$ is satisfied, since $x + 1 = 8 > 0$ .

Solution: $x = 7$ .

The base of the logarithm is $\frac{1}{2}$ and the method does not change: by the definition of logarithm the argument equals $(\frac{1}{2})^{- 3}$ . The negative exponent replaces the fraction with its reciprocal, so $(\frac{1}{2})^{- 3} = 2^{3} = 8$ . From the equation $x + 1 = 8$ it follows that $x = 7$ ; the check confirms the result, because $lo g_{1/2} 8 = - 3$ since $(\frac{1}{2})^{- 3} = 8$ .