If x is a real number greater than 2 such that \log_2(x^2 - 4) - \log_2(x - 2) = 3, what i… — ACT

If $x$ is a real number greater than 2 such that $\log_2(x^2 - 4) - \log_2(x - 2) = 3$ , what is the value of $x$ ?

The value of

x

is 6.

Applying the quotient property of logarithms simplifies the equation to

\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3

. Factoring the numerator as

(x-2)(x+2)

and canceling the common factor

(x-2)

yields

\log_2(x+2) = 3

. Rewriting this in exponential form gives

x + 2 = 2^3 = 8

, which yields

x = 6

Combine the logarithmic terms on the left side of the equation using the quotient property of logarithms:

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

\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3

Consolidating the two logarithmic terms simplifies the equation into a single logarithmic term.

Factor the numerator as a difference of squares and simplify the rational expression.

\log_2(x + 2) = 3

(since

x > 2

x - 2 \neq 0

)

Factoring

x^2 - 4

(x - 2)(x + 2)

allows the cancellation of the common factor

(x - 2)

Convert the logarithmic equation into its equivalent exponential form:

\log_b Y = c \Rightarrow Y = b^c

x + 2 = 2^3

This removes the logarithm and sets up a linear equation to solve.

Evaluate the exponent and solve for

x

x + 2 = 8 \Rightarrow x = 6

Evaluating

2^3

as 8 and subtracting 2 from both sides isolates

x

Solving logarithmic equations using properties of logarithms and algebraic factoring

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