Math Problem Statement

Solve the equation: 4log_x 3 + 2log_9 x = 5 where x = 5

Solution

The equation shown in the image is:

4log⁡x3+2log⁡9x=54 \log_x 3 + 2 \log_9 x = 5

Here, log⁡x3\log_x 3 represents the logarithm of 3 with base xx, and log⁡9x\log_9 x represents the logarithm of xx with base 9. The goal is to solve for xx, given the equation.

Step-by-step solution:

1. Use change of base formula:

For log⁡x3\log_x 3 and log⁡9x\log_9 x, we apply the change of base formula, which is: log⁡ba=log⁡alog⁡b\log_b a = \frac{\log a}{\log b} This gives us:

\quad \text{and} \quad \log_9 x = \frac{\log x}{\log 9}$$ Thus, the original equation becomes: $$4 \times \frac{\log 3}{\log x} + 2 \times \frac{\log x}{\log 9} = 5$$ #### 2. Simplify the equation: We know that $$\log 9 = 2 \log 3$$, since $$9 = 3^2$$, so we substitute $$\log 9$$: $$4 \times \frac{\log 3}{\log x} + 2 \times \frac{\log x}{2 \log 3} = 5$$ Simplifying: $$4 \times \frac{\log 3}{\log x} + \frac{\log x}{\log 3} = 5$$ #### 3. Set substitution: Let $$y = \log x / \log 3$$, so the equation becomes: $$4 \times \frac{1}{y} + y = 5$$ Multiplying through by $$y$$ to eliminate the denominator: $$4 + y^2 = 5y$$ Rearrange this into a standard quadratic form: $$y^2 - 5y + 4 = 0$$ #### 4. Solve the quadratic equation: Using the quadratic formula: $$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$$ $$y = \frac{5 \pm \sqrt{25 - 16}}{2}$$ $$y = \frac{5 \pm \sqrt{9}}{2}$$ $$y = \frac{5 \pm 3}{2}$$ Thus, the two solutions for $$y$$ are: $$y = \frac{5 + 3}{2} = 4 \quad \text{or} \quad y = \frac{5 - 3}{2} = 1$$ #### 5. Back-substitute for $$y$$: Recall that $$y = \log x / \log 3$$, so: $$\frac{\log x}{\log 3} = 4 \quad \text{or} \quad \frac{\log x}{\log 3} = 1$$ This leads to: $$\log x = 4 \log 3 \quad \text{or} \quad \log x = \log 3$$ #### 6. Solve for $$x$$: From $$\log x = 4 \log 3$$, we have: $$x = 3^4 = 81$$ From $$\log x = \log 3$$, we have: $$x = 3$$ Thus, the two possible values of $$x$$ are $$x = 81$$ and $$x = 3$$. #### 7. Verify the solutions: Plugging $$x = 81$$ and $$x = 3$$ back into the original equation: - For $$x = 3$$: The left side of the equation becomes undefined due to $$\log_3 3$$ (since the equation would involve $$\log_3 3 = 1$$, leading to discrepancies). - For $$x = 81$$: The equation holds. Thus, the correct solution is $$x = 81$$. --- Would you like further details or have any questions on this process? Here are 5 related questions to explore: 1. What is the change of base formula for logarithms? 2. How does the quadratic formula solve quadratic equations? 3. Why is $$\log_9 x$$ equivalent to $$\frac{\log x}{\log 9}$$? 4. What are the properties of logarithms and how do they simplify expressions? 5. How can logarithmic equations be used in real-world applications? **Tip:** When solving logarithmic equations, always check for extraneous solutions, as not all results may satisfy the original equation.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Logarithms
Change of Base Formula
Quadratic Equations

Formulas

Change of base formula: log_b(a) = log(a) / log(b)
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Change of Base Theorem
Quadratic Formula

Suitable Grade Level

Grades 10-12

Related Recommendation

Solve Logarithmic Equation log_3 (x^3) - 4log_9 (x) - 5 log_27 (sqrt(x)) = log_9 (4)

Solve x^(log3(x)) = (1/9)x^3 Using Logarithms and Quadratic Formula

Solve the Logarithmic Equation 5log_5(x) + log_x(5) = 9

Solve the Logarithmic Equation log_12(4^(3x) + 3^x - 9)

Solving Logarithmic Equations: log10(x) + log10(x+9) = 1, log2(x-1) - log2(x-5) = 3