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Zorluk: OrtaIndices and Logarithms

Find the value of xx that satisfies the logarithmic equation log⁡2(x2−1)−log⁡2(x−1)=3\log_2(x^2 - 1) - \log_2(x - 1) = 3.

Cevap: 7

Cevap

The value of xx is 77.
Applying the logarithm quotient law reduces log⁡2(x2−1)−log⁡2(x−1)\log_2(x^2 - 1) - \log_2(x - 1) to log⁡2(x2−1x−1)\log_2\left(\frac{x^2 - 1}{x - 1}\right). Factoring x2−1x^2 - 1 into (x−1)(x+1)(x - 1)(x + 1) and canceling (x−1)(x - 1) leaves log⁡2(x+1)=3\log_2(x + 1) = 3. Converting this to exponential form gives x+1=23=8x + 1 = 2^3 = 8, which yields x=7x = 7.

Adım Adım Çözüm

1
Apply the quotient rule for logarithms to combine the left-hand side into a single logarithmic expression.
log⁡2(x2−1x−1)=3\log_2\left(\frac{x^2 - 1}{x - 1}\right) = 3
According to logarithmic rules, the difference of two logarithms of the same base is equal to the logarithm of their quotient: log⁡bA−log⁡bB=log⁡b(AB)\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right).
2
Factor the numerator as a difference of two squares and simplify the rational expression.
log⁡2(x+1)=3\log_2(x + 1) = 3
Since x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1), dividing by (x−1)(x - 1) leaves (x+1)(x + 1), given x≠1x \neq 1.
3
Rewrite the logarithmic equation in exponential form.
x+1=23=8x + 1 = 2^3 = 8
By definition of logarithms, log⁡by=z  ⟺  bz=y\log_b y = z \iff b^z = y.
4
Solve the resulting linear equation for xx.
x=7x = 7
Subtracting 11 from both sides gives x=7x = 7.

Anahtar Kavram

Quotient Law of Logarithms and Logarithmic Equations
Tahmini Süre:1m 30s
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Indices and Logarithms

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