The equation in the image is:

$x^{\log_3 x} = \frac{1}{9}x^3$

Let me solve this step by step.

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Step 1: Simplify the equation

The equation can be rewritten as:

$x^{\log_3 x} = \frac{1}{9} x^3$

We know that $\frac{1}{9} = 3^{-2}$, so:

$x^{\log_3 x} = 3^{-2} x^3$

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Step 2: Take the logarithm of both sides

To simplify the left-hand side, let’s take the natural logarithm ($\ln$) of both sides:

$\ln(x^{\log_3 x}) = \ln(3^{-2} x^3)$

Using the logarithmic property $\ln(a^b) = b \ln(a)$, this becomes:

$(\log_3 x) \ln(x) = \ln(3^{-2}) + \ln(x^3)$

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Step 3: Express $\log_3 x$ in terms of natural logarithms

Recall that $\log_3 x = \frac{\ln(x)}{\ln(3)}$. Substituting this:

$\frac{\ln(x)}{\ln(3)} \ln(x) = \ln(3^{-2}) + \ln(x^3)$

$\frac{\ln^2(x)}{\ln(3)} = -2 \ln(3) + 3 \ln(x)$

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Step 4: Rearrange the equation

Multiply through by $\ln(3)$ to eliminate the denominator:

$\ln^2(x) = -2 \ln^2(3) + 3 \ln(3) \ln(x)$

Reorganizing:

$\ln^2(x) - 3 \ln(3) \ln(x) + 2 \ln^2(3) = 0$

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Step 5: Solve the quadratic equation

This is a quadratic equation in terms of $\ln(x)$. Let $y = \ln(x)$, so the equation becomes:

$y^2 - 3 \ln(3) y + 2 \ln^2(3) = 0$

Solve this quadratic equation using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here:

• $a = 1$,
• $b = -3 \ln(3)$,
• $c = 2 \ln^2(3)$.

Substitute:

$y = \frac{-(-3 \ln(3)) \pm \sqrt{(-3 \ln(3))^2 - 4(1)(2 \ln^2(3))}}{2(1)}$

$y = \frac{3 \ln(3) \pm \sqrt{9 \ln^2(3) - 8 \ln^2(3)}}{2}$

$y = \frac{3 \ln(3) \pm \sqrt{\ln^2(3)}}{2}$

$y = \frac{3 \ln(3) \pm \ln(3)}{2}$

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Step 6: Find the two solutions

Split into two cases:

1. $y = \frac{3 \ln(3) + \ln(3)}{2} = \frac{4 \ln(3)}{2} = 2 \ln(3)$
2. $y = \frac{3 \ln(3) - \ln(3)}{2} = \frac{2 \ln(3)}{2} = \ln(3)$

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Step 7: Convert back to $x$

Recall that $y = \ln(x)$, so:

1. $\ln(x) = 2 \ln(3) \implies x = e^{2 \ln(3)} = 3^2 = 9$
2. $\ln(x) = \ln(3) \implies x = e^{\ln(3)} = 3$

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Final Answer:

The solutions are:

$x = 3 \quad \text{and} \quad x = 9$

Would you like further details or clarification? 😊

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Related Questions:

1. How do you simplify logarithmic equations like this one?
2. What are the properties of logarithms and exponents that are helpful for solving equations?
3. How do you use the quadratic formula in different contexts?
4. Why do we take logarithms on both sides of an equation in such problems?
5. What are the key steps for rewriting equations involving powers and logarithms?

Tip:

When working with equations involving logarithms and exponents, always look for ways to simplify using properties like $\log(a^b) = b \log(a)$ or $\log(ab) = \log(a) + \log(b)$.