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Zorluk: OrtaIndices and Logarithms

What is the positive integer solution to the logarithmic equation log2x3logx2=2\log_2 x - 3\log_x 2 = 2?

  1. 88Cevap
  2. B
    66
  3. C
    33
  4. D
    1212

Cevap

The positive integer solution is 88.
Using the change of base identity logx2=1log2x\log_x 2 = \frac{1}{\log_2 x}, we rewrite the equation as u3u=2u - \frac{3}{u} = 2 where u=log2xu = \log_2 x. Rearranging yields the quadratic u22u3=0u^2 - 2u - 3 = 0, which factors as (u3)(u+1)=0(u - 3)(u + 1) = 0. This gives u=3u = 3 or u=1u = -1. Converting back to x=2ux = 2^u, we get x=23=8x = 2^3 = 8 or x=21=12x = 2^{-1} = \frac{1}{2}. Since the question asks for the positive integer solution, the correct answer is 8.

Adım Adım Çözüm

1
Apply the change of base rule to express logx2\log_x 2 in terms of base 2.
log2x3log2x=2\log_2 x - \frac{3}{\log_2 x} = 2
The change of base identity states that logba=1logab\log_b a = \frac{1}{\log_a b}.
2
Substitute u=log2xu = \log_2 x into the equation and clear the fraction.
u3u=2    u22u3=0u - \frac{3}{u} = 2 \implies u^2 - 2u - 3 = 0
Multiplying through by uu (where u0u \neq 0) transforms the equation into standard quadratic form.
3
Factor the quadratic equation to solve for uu.
(u3)(u+1)=0    u=3 or u=1(u - 3)(u + 1) = 0 \implies u = 3 \text{ or } u = -1
Finding the roots of the quadratic equation in terms of uu.
4
Convert back to xx using x=2ux = 2^u and select the positive integer solution.
For u=3u = 3: x=23=8x = 2^3 = 8. For u=1u = -1: x=21=12x = 2^{-1} = \frac{1}{2}.
The question asks specifically for the positive integer solution, which is 8.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction
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