Find the real value of x that satisfies the logarithmic equation \log_2(x^2 + 3x - 22) - \… — JAMB UTME

Find the real value of $x$ that satisfies the logarithmic equation $\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3$ .

Answer: 6

The real value of

x

that satisfies the equation is

6

Applying the logarithmic quotient rule transforms

\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3

into

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

. Expressing this in exponential form yields

\frac{x^2 + 3x - 22}{x - 2} = 8

, which simplifies to

x^2 - 5x - 6 = 0

. Factoring gives

(x - 6)(x + 1) = 0

. Since the argument of a logarithm must be strictly positive (

x - 2 > 0 \implies x > 2

x = -1

is extraneous and

x = 6

is the only valid solution.

Combine the logarithmic terms using the quotient law of logarithms.

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

The difference of two logarithms with the same base equals the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

Convert the logarithmic equation into its equivalent exponential form.

\frac{x^2 + 3x - 22}{x - 2} = 2^3 = 8

\log_b Y = c

, then

Y = b^c

Clear the denominator and simplify to form a quadratic equation.

x^2 - 5x - 6 = 0

Multiplying both sides by

(x - 2)

yields

x^2 + 3x - 22 = 8x - 16

, which rearranges to

x^2 - 5x - 6 = 0

Solve the quadratic equation by factoring.

(x - 6)(x + 1) = 0 \implies x = 6 \text{ or } x = -1

The factors of

-6

that sum to

-5

are

-6

and

1

Test roots against domain restrictions to eliminate extraneous solutions.

x = 6

Logarithmic arguments must be strictly positive. For

\log_2(x - 2)

to be defined,

x > 2

. Thus,

x = -1

is extraneous, leaving

x = 6

as the unique valid solution.

Logarithmic Equations and Domain Restrictions

Question