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Difficulty: HardLogarithmic and Exponential Expressions and Equations

What is the product of all real values of xx that satisfy the equation log⁡3(x)−6log⁡x(3)=1\log_3(x) - 6\log_x(3) = 1?

Answer: 3

Answer

The product of all real values of xx that satisfy the equation is 3.
By applying the change-of-base formula, the equation becomes log⁡3(x)−6log⁡3(x)=1\log_3(x) - \frac{6}{\log_3(x)} = 1. Substituting y=log⁡3(x)y = \log_3(x) leads to y2−y−6=0y^2 - y - 6 = 0, which has roots y=3y = 3 and y=−2y = -2. These roots correspond to x=33=27x = 3^3 = 27 and x=3−2=19x = 3^{-2} = \frac{1}{9}. Both solutions are valid because they are positive and do not equal 1. The product of these solutions is 27×19=327 \times \frac{1}{9} = 3. Alternatively, using Vieta's formulas, the sum of the roots of the quadratic equation is y1+y2=1y_1 + y_2 = 1. The product of the solutions is x1x2=3y1⋅3y2=3y1+y2=31=3x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3.

Step-by-Step Solution

1
Apply the change-of-base formula to rewrite the variable base term.
log⁡3(x)−6log⁡3(x)=1\log_3(x) - \frac{6}{\log_3(x)} = 1
This expresses the equation in terms of logarithms with the same base.
2
Use substitution to convert the equation into a quadratic form.
y−6y=1⇒y2−y−6=0y - \frac{6}{y} = 1 \Rightarrow y^2 - y - 6 = 0 where y=log⁡3(x)y = \log_3(x)
Substitution simplifies the logarithmic equation into a polynomial equation.
3
Solve the quadratic equation by factoring.
(y−3)(y+2)=0⇒y=3(y-3)(y+2) = 0 \Rightarrow y = 3 or y=−2y = -2
Finding the roots for yy is the intermediate step to solving for xx.
4
Back-substitute to find the values of xx.
x=33=27x = 3^3 = 27 and x=3−2=19x = 3^{-2} = \frac{1}{9}
Converting from logarithmic form back to exponential form yields the values of xx.
5
Multiply the solutions together.
27×19=327 \times \frac{1}{9} = 3
The question asks for the product of all real solutions.

Key Concept

Solving logarithmic equations using the change-of-base formula and quadratic substitution.

Alternative Method

Instead of solving for individual values of xx, note that if y1y_1 and y2y_2 are the roots of the quadratic equation y2−y−6=0y^2 - y - 6 = 0, then y1+y2=1y_1 + y_2 = 1 by Vieta's formulas. Since x1=3y1x_1 = 3^{y_1} and x2=3y2x_2 = 3^{y_2}, the product of the solutions is x1x2=3y1⋅3y2=3y1+y2=31=3x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3.
Estimated Time:2m 0s
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Topics

Logarithmic and Exponential Expressions and Equations

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