Worked examples · Sheet 2

Exponential equations solved with the logarithm

Theory used: Exponential equation with a constant right-hand side Equality of powers with the same base

Exercise 1

Solve the equation

3^{x} = 7.

3^{x} = 7 ⟺ x = lo g_{3} 7

Solution: $x = lo g_{3} 7$ .

The number $7$ is not a power of $3$ with a recognizable rational exponent, so the equation cannot be reduced to the equality of powers with the same base in elementary form. By definition, the logarithm $lo g_{3} 7$ is the exponent to which $3$ must be raised to obtain $7$ , therefore the solution is $x = lo g_{3} 7$ . The right-hand side is positive, so $lo g_{3} 7$ is defined and the solution exists; it is unique because the exponential function is injective.

Exercise 2

Solve the equation

2^{x - 3} = 5.

2^{x - 3} = 5 ⟺ x - 3 = lo g_{2} 5 ⟺ x = 3 + lo g_{2} 5

Solution: $x = 3 + lo g_{2} 5$ .

The equation has the form $2^{A} = 5$ with $A = x - 3$ , so by the definition of logarithm the exponent equals $A = lo g_{2} 5$ . What remains is the first-degree equation $x - 3 = lo g_{2} 5$ , from which $x = 3 + lo g_{2} 5$ ; the value $lo g_{2} 5$ is treated as a number, which lies between $2$ and $3$ because $2^{2} = 4 < 5 < 8 = 2^{3}$ .

Exercise 3

Solve the equation

5^{2 x + 1} = 2.

5^{2 x + 1} = 2 ⟺ 2 x + 1 = lo g_{5} 2 ⟺ x = \frac{lo g _{5} 2 - 1}{2}

Solution: $x = \frac{lo g _{5} 2 - 1}{2}$ .

The number $2$ is not a power of $5$ with a recognizable exponent, so by the definition of logarithm the whole exponent equals $2 x + 1 = lo g_{5} 2$ . The first-degree equation is then solved: subtracting $1$ and dividing by $2$ gives $x = \frac{lo g _{5} 2 - 1}{2}$ . The result is negative, because $2 < 5 = 5^{1}$ and the base $5$ is greater than $1$ , so the exponent $lo g_{5} 2$ is less than $1$ and the numerator $lo g_{5} 2 - 1$ is negative.