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Zorluk: OrtaLogarithmic and Exponential Expressions and Equations

If xx is a real number greater than 2 such that log⁡2(x2−4)−log⁡2(x−2)=3\log_2(x^2 - 4) - \log_2(x - 2) = 3, what is the value of xx?

  1. A
    1
  2. B
    4
  3. 6Cevap
  4. D
    7
  5. E
    16

Cevap

The value of xx is 6.
Applying the quotient property of logarithms simplifies the equation to log⁡2(x2−4x−2)=3\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3. Factoring the numerator as (x−2)(x+2)(x-2)(x+2) and canceling the common factor (x−2)(x-2) yields log⁡2(x+2)=3\log_2(x+2) = 3. Rewriting this in exponential form gives x+2=23=8x + 2 = 2^3 = 8, which yields x=6x = 6.

Adım Adım Çözüm

1
Combine the logarithmic terms on the left side of the equation using the quotient property of logarithms: log⁡bA−log⁡bB=log⁡b(AB)\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right).
log⁡2(x2−4x−2)=3\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3
Consolidating the two logarithmic terms simplifies the equation into a single logarithmic term.
2
Factor the numerator as a difference of squares and simplify the rational expression.
log⁡2(x+2)=3\log_2(x + 2) = 3 (since x>2x > 2, x−2≠0x - 2 \neq 0)
Factoring x2−4x^2 - 4 as (x−2)(x+2)(x - 2)(x + 2) allows the cancellation of the common factor (x−2)(x - 2).
3
Convert the logarithmic equation into its equivalent exponential form: log⁡bY=c⇒Y=bc\log_b Y = c \Rightarrow Y = b^c.
x+2=23x + 2 = 2^3
This removes the logarithm and sets up a linear equation to solve.
4
Evaluate the exponent and solve for xx.
x+2=8⇒x=6x + 2 = 8 \Rightarrow x = 6
Evaluating 232^3 as 8 and subtracting 2 from both sides isolates xx.

Anahtar Kavram

Solving logarithmic equations using properties of logarithms and algebraic factoring
Tahmini Süre:1m 30s
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