If \log_2(x + 5) + \log_2(x - 1) = 4, what is the value of x? — ACT

If $\log_2(x + 5) + \log_2(x - 1) = 4$ , what is the value of $x$ ?

Answer: 3

The correct answer is 3.

The correct answer is 3. Combining the logarithmic terms using the product property gives

\log_2((x + 5)(x - 1)) = 4

. Converting this to exponential form yields

(x + 5)(x - 1) = 2^4 = 16

. Expanding and rewriting in standard form gives

x^2 + 4x - 21 = 0

, which factors as

(x + 7)(x - 3) = 0

. This gives potential solutions of

x = 3

and

x = -7

. However,

x = -7

results in negative arguments for the logarithms in the original equation, making it extraneous. Thus, the only valid solution is 3.

Apply the product property of logarithms to combine the terms on the left side of the equation.

\log_2((x + 5)(x - 1)) = 4

The sum of logarithms with the same base is equal to the logarithm of their product:

\log_b(A) + \log_b(B) = \log_b(AB)

Convert the logarithmic equation to its equivalent exponential form.

(x + 5)(x - 1) = 2^4

By definition,

\log_b(Y) = C

is equivalent to

b^C = Y

Expand the product and simplify the equation into standard quadratic form.

x^2 + 4x - 21 = 0

Expanding

(x + 5)(x - 1)

yields

x^2 + 4x - 5

. Subtracting 16 from both sides gives the standard quadratic form

Ax^2 + Bx + C = 0

Factor the quadratic equation to find the potential values of

x

(x + 7)(x - 3) = 0

, so

x = -7

x = 3

We need two numbers that multiply to

-21

and add to

4

, which are

7

and

-3

Check the potential solutions against the domain of the original logarithmic equation.

x = 3

The arguments of the logarithms,

x + 5

and

x - 1

, must be strictly positive. For

x = -7

, the arguments are negative, so

x = -7

is an extraneous solution. For

x = 3

, both arguments are positive, so

x = 3

is the only valid solution.

Solving logarithmic equations by applying logarithmic properties and converting to exponential form, while checking for extraneous solutions.

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