Solve for x in the logarithmic equation \log_2 (x + 2) + \log_2 (x - 4) = 4. — JAMB UTME

Solve for $x$ in the logarithmic equation $\log_2 (x + 2) + \log_2 (x - 4) = 4$ .

The value of

x

is 6.

Using the product rule for logarithms,

\log_2 (x + 2) + \log_2 (x - 4) = \log_2 [(x + 2)(x - 4)] = 4

. Converting to exponential form yields

(x + 2)(x - 4) = 2^4 = 16

. Expanding gives

x^2 - 2x - 8 = 16

, which simplifies to

x^2 - 2x - 24 = 0

. Factoring gives

(x - 6)(x + 4) = 0

. Since the logarithmic arguments must be positive (

x > 4

), the negative root

-4

is discarded, leaving

x = 6

Apply the product rule of logarithms:

\log_b M + \log_b N = \log_b (M \cdot N)

\log_2 [(x + 2)(x - 4)] = 4

Logarithms with the same base being added combine by multiplying their arguments.

Convert the logarithmic equation to its exponential form

(x + 2)(x - 4) = 2^4 = 16

\log_b Y = X

, then

Y = b^X

Expand and rearrange into a standard quadratic equation

x^2 - 2x - 8 = 16 \implies x^2 - 2x - 24 = 0

Expanding

(x+2)(x-4)

yields

x^2 - 2x - 8

, and subtracting 16 sets the equation to zero.

Factor the quadratic equation to find potential solutions

(x - 6)(x + 4) = 0 \implies x = 6 \text{ or } x = -4

The roots of

x^2 - 2x - 24 = 0

are

x = 6

and

x = -4

Check the domain restrictions for logarithmic functions

x = 6

The arguments of the original logarithms require

x + 2 > 0 \implies x > -2

and

x - 4 > 0 \implies x > 4

. Therefore,

x = -4

is extraneous and

x = 6

is the only valid solution.

Solving logarithmic equations using product rule and domain constraints

Question