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Difficulty: Very hardLogarithms and Change of Base

If log⁡3x−2log⁡x3=1\log_3 x - 2 \log_x 3 = 1, what is the sum of all valid solutions for xx?

  1. 283\frac{28}{3}Answer
  2. B
    1212
  3. C
    11
  4. D
    99

Answer

The sum of all valid solutions for xx is 283\frac{28}{3}.
By using the change of base identity log⁡x3=1log⁡3x\log_x 3 = \frac{1}{\log_3 x} and substituting u=log⁡3xu = \log_3 x, the given equation simplifies to u−2u=1u - \frac{2}{u} = 1. Multiplying by uu gives the quadratic equation u2−u−2=0u^2 - u - 2 = 0, which factors as (u−2)(u+1)=0(u - 2)(u + 1) = 0. This yields u=2u = 2 or u=−1u = -1. Converting back to xx gives x=32=9x = 3^2 = 9 and x=3−1=13x = 3^{-1} = \frac{1}{3}. Both values satisfy domain constraints for logarithmic base (x>0x > 0 and x≠1x \neq 1). Adding these values gives 9+13=2839 + \frac{1}{3} = \frac{28}{3}.

Step-by-Step Solution

1
Apply the change of base formula to express the equation with a common logarithmic base
Using log⁡x3=1log⁡3x\log_x 3 = \frac{1}{\log_3 x}, the equation becomes log⁡3x−2log⁡3x=1\log_3 x - \frac{2}{\log_3 x} = 1.
Change of base allows all terms containing xx to be represented in terms of a single logarithmic expression.
2
Substitute u=log⁡3xu = \log_3 x and clear the fraction to form a quadratic equation
u−2u=1  ⟹  u2−u−2=0u - \frac{2}{u} = 1 \implies u^2 - u - 2 = 0.
Multiplying through by uu (where u≠0u \neq 0) transforms the logarithmic relationship into a standard quadratic algebraic equation.
3
Factor the quadratic equation to find the values of uu
(u−2)(u+1)=0  ⟹  u=2 or u=−1(u - 2)(u + 1) = 0 \implies u = 2 \text{ or } u = -1.
Factoring isolates the linear roots for the substituted variable uu.
4
Solve for xx from u=log⁡3xu = \log_3 x and verify validity within the domain
For u=2u = 2: log⁡3x=2  ⟹  x=32=9\log_3 x = 2 \implies x = 3^2 = 9.
For u=−1u = -1: log⁡3x=−1  ⟹  x=3−1=13\log_3 x = -1 \implies x = 3^{-1} = \frac{1}{3}.
Both solutions are positive and x≠1x \neq 1, so both are valid.
Converting from logarithmic form to exponential form retrieves the original variable xx.
5
Calculate the sum of all valid solutions
Sum = 9+13=273+13=2839 + \frac{1}{3} = \frac{27}{3} + \frac{1}{3} = \frac{28}{3}.
The question requires finding the total sum of all permissible real solutions for xx.

Key Concept

Logarithmic Change of Base and Equations Reducible to Quadratics
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Topics

Logarithms and Change of Base

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