$\log _{3} x-4 \log _{x} 3=1+\frac{1}{\log _{9} 3}$

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Precalculus

Quotient property of logarithms

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\log _{3} x-4 \log _{x} 3=1+\frac{1}{\log _{9} 3}

Understand given logarithmic equation: Understand the given logarithmic equation. $\newline$ We have the equation $\log_{3}x - 4\log_{x}3 = 1 + \frac{1}{\log_{9}3}$ . We need to solve for $x$ .
Use change of base formula: Use the change of base formula to rewrite $4\log_{x}3$ . Change of base formula: $\log_{a}b = \frac{\log(c)b}{\log(c)a}$ $4\log_{x}3 = 4 \times \left(\frac{\log(3)}{\log(x)}\right)$
Simplify right side: Simplify the right side of the equation. $\newline$ We have $(1)/(\log_{9}3)$ . Using the change of base formula again, we get: $\newline$ $(1)/(\log_{9}3) = \log(3) / \log(9)$ $\newline$ Since $9$ is $3^2$ , $\log(9) = \log(3^2) = 2\log(3)$ , so: $\newline$ $(1)/(\log_{9}3) = \log(3) / (2\log(3)) = 1/2$
Substitute simplified expressions: Substitute the simplified expressions back into the equation. $\newline$ Now we have $\log_{3}x - 4 \times \left(\frac{\log(3)}{\log(x)}\right) = 1 + \frac{1}{2}$
Combine constants on right side: Combine the constants on the right side of the equation. $\newline$ $1 + \frac{1}{2} = \frac{3}{2}$ $\newline$ So, the equation becomes $\log_{3}x - 4 \cdot \left(\frac{\log(3)}{\log(x)}\right) = \frac{3}{2}$
Convert to natural logarithm: Convert $\log_{3}x$ to the natural logarithm using the change of base formula. $\newline$ $\log_{3}x = \frac{\log(x)}{\log(3)}$ $\newline$ The equation now is $\left(\frac{\log(x)}{\log(3)}\right) - 4 \cdot \left(\frac{\log(3)}{\log(x)}\right) = \frac{3}{2}$
Multiply through to clear denominators: Multiply through by $\log(3)\log(x)$ to clear the denominators. $\$\log(x) \cdot \log(x)$ - $4$ \cdot (\log( $3$ ) \cdot \log( $3$ )) = \left(\frac{ $3$ }{ $2$ }\right) \cdot \log( $3$ ) \cdot \log(x)\)
Simplify the equation: Simplify the equation. $\newline$ $(\log(x))^2 - 4 \cdot (\log(3))^2 = \frac{3}{2} \cdot \log(3) \cdot \log(x)$
Rearrange into quadratic form: Rearrange the equation into a quadratic form. $\newline$ Let's set $A = \log(x)$ and $B = \log(3)$ , then we have: $\newline$ $A^2 - \frac{3}{2} \cdot B \cdot A - 4 \cdot B^2 = 0$
Solve quadratic equation for A: Solve the quadratic equation for A. $\newline$ This is a quadratic equation in the form of $A^2 - C \cdot A - D = 0$ , where $C = \left(\frac{3}{2}\right) \cdot B$ and $D = 4 \cdot B^2$ . $\newline$ We can use the quadratic formula to solve for A: $A = \frac{C \pm \sqrt{C^2 + 4D}}{2}$
Calculate discriminant: Calculate the discriminant of the quadratic equation. $\newline$ The discriminant is $C^2 + 4D = \left(\frac{3}{2} \cdot B\right)^2 + 4 \cdot 4 \cdot B^2$
Substitute $B$ into discriminant: Substitute $B = \log(3)$ into the discriminant. $\newline$ Discriminant = $\left(\frac{3}{2} \cdot \log(3)\right)^2 + 4 \cdot 4 \cdot (\log(3))^2$
Simplify the discriminant: Simplify the discriminant. $\newline$ Discriminant = $(\frac{9}{4}) \cdot (\log(3))^2 + 16 \cdot (\log(3))^2$ $\newline$ Discriminant = $(\frac{9}{4} + 16) \cdot (\log(3))^2$ $\newline$ Discriminant = $(\frac{9}{4} + \frac{64}{4}) \cdot (\log(3))^2$ $\newline$ Discriminant = (\frac{$73$}{$4$}) \cdot (\log(\(3))^ $2$
Solve for A using quadratic formula: Solve for A using the quadratic formula. $\newline$ $A = \frac{\left(\frac{3}{2}\right) * B \pm \sqrt{\left(\frac{73}{4}\right) * B^2}}{2}$ $\newline$ $A = \frac{\left(\frac{3}{2}\right) * \log(3) \pm \sqrt{\left(\frac{73}{4}\right) * (\log(3))^2}}{2}$
Solve for x: Since $A = \log(x)$ , we can solve for x. $x = 10^A$ $x = 10^{\left(\frac{3}{2} \cdot \log(3) \pm \sqrt{\frac{73}{4} \cdot (\log(3))^2}\right) / 2}$
Realize mistake in previous step: Realize there is a mistake in the previous step. $\newline$ The base of the logarithm is $3$ , not $10$ . Therefore, we should use $3$ as the base for exponentiation, not $10$ . $\newline$ $x = 3^A$ $\newline$ $x = 3^{\left(\frac{3}{2} \cdot \log(3) \pm \sqrt{\frac{73}{4} \cdot (\log(3))^2}\right) / 2}$