Math Problem Statement

5log_5(x) + log_x(5) = 9

Solution

The equation provided is:

5log5(x)+logx(5)=95 \log_5(x) + \log_x(5) = 9

Let's solve this step-by-step:


Step 1: Simplify log5(x)\log_5(x)

Using the logarithmic property logb(a)=log(a)log(b)\log_b(a) = \frac{\log(a)}{\log(b)}, we write: log5(x)=log(x)log(5).\log_5(x) = \frac{\log(x)}{\log(5)}.


Step 2: Simplify logx(5)\log_x(5)

Similarly, for logx(5)\log_x(5), we write: logx(5)=log(5)log(x).\log_x(5) = \frac{\log(5)}{\log(x)}.


Step 3: Substitute these into the equation

Substituting these into the original equation: 5log(x)log(5)+log(5)log(x)=9.5 \cdot \frac{\log(x)}{\log(5)} + \frac{\log(5)}{\log(x)} = 9.

Let y=log(x)y = \log(x), so the equation becomes: 5ylog(5)+log(5)y=9.5 \cdot \frac{y}{\log(5)} + \frac{\log(5)}{y} = 9.


Step 4: Eliminate the fraction

Multiply through by ylog(5)y \cdot \log(5) to eliminate the denominators: 5y2+(log(5))2=9ylog(5).5y^2 + (\log(5))^2 = 9y \cdot \log(5).


Step 5: Rearrange into a standard quadratic form

Reorganize the terms: 5y29ylog(5)+(log(5))2=0.5y^2 - 9y \cdot \log(5) + (\log(5))^2 = 0.

This is a quadratic equation in yy, where:

  • a=5a = 5,
  • b=9log(5)b = -9 \cdot \log(5),
  • c=(log(5))2c = (\log(5))^2.

Step 6: Solve using the quadratic formula

The quadratic formula is: y=b±b24ac2a.y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Substituting aa, bb, and cc: y=9log(5)±(9log(5))245(log(5))225.y = \frac{9 \cdot \log(5) \pm \sqrt{(9 \cdot \log(5))^2 - 4 \cdot 5 \cdot (\log(5))^2}}{2 \cdot 5}.

Simplify: y=9log(5)±81(log(5))220(log(5))210.y = \frac{9 \cdot \log(5) \pm \sqrt{81 \cdot (\log(5))^2 - 20 \cdot (\log(5))^2}}{10}.

y=9log(5)±61(log(5))210.y = \frac{9 \cdot \log(5) \pm \sqrt{61 \cdot (\log(5))^2}}{10}.

y=9log(5)±log(5)6110.y = \frac{9 \cdot \log(5) \pm \log(5) \cdot \sqrt{61}}{10}.

Factor out log(5)\log(5): y=log(5)9±6110.y = \log(5) \cdot \frac{9 \pm \sqrt{61}}{10}.


Step 7: Back-substitute for y=log(x)y = \log(x)

log(x)=log(5)9±6110.\log(x) = \log(5) \cdot \frac{9 \pm \sqrt{61}}{10}.

Exponentiate both sides to solve for xx: x=59±6110.x = 5^{\frac{9 \pm \sqrt{61}}{10}}.

Thus, there are two solutions: x=59+6110andx=596110.x = 5^{\frac{9 + \sqrt{61}}{10}} \quad \text{and} \quad x = 5^{\frac{9 - \sqrt{61}}{10}}.


Final Answer:

The solutions are: x=59+6110,x=596110.x = 5^{\frac{9 + \sqrt{61}}{10}}, \quad x = 5^{\frac{9 - \sqrt{61}}{10}}.

Would you like a deeper explanation of any step?


Here are 5 related questions to explore:

  1. How can the quadratic formula be used in logarithmic equations more generally?
  2. What are the properties of logarithms and how are they applied in problem-solving?
  3. How do changes of base for logarithms affect their use in equations?
  4. What happens if logx(5)\log_x(5) is written directly as its inverse (1/log5(x))(1/\log_5(x))?
  5. Can these solutions for xx be expressed in terms of decimals instead of logarithmic expressions?

Tip: Always check if logarithmic solutions satisfy the domain of the original equation (e.g., x>0,x1x > 0, x \neq 1).

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Math Problem Analysis

Mathematical Concepts

Logarithms
Quadratic Equations
Change of Base Formula

Formulas

log_b(a) = log(a) / log(b)
Quadratic formula: y = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

Properties of Logarithms
Change of Base Formula
Quadratic Equation Solving

Suitable Grade Level

Grades 11-12