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log_(3)x-4log_(x)3=1+(1)/(log_(9)3)

log⁡3x−4log⁡x3=1+1log⁡93 \log _{3} x-4 \log _{x} 3=1+\frac{1}{\log _{9} 3}

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Quotient property of logarithmsright chevron icon

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Q. log⁡3x−4log⁡x3=1+1log⁡93 \log _{3} x-4 \log _{x} 3=1+\frac{1}{\log _{9} 3}
  1. Understand given logarithmic equation: Understand the given logarithmic equation.\newlineWe have the equation log⁡3x−4log⁡x3=1+1log⁡93\log_{3}x - 4\log_{x}3 = 1 + \frac{1}{\log_{9}3}. We need to solve for xx.
  2. Use change of base formula: Use the change of base formula to rewrite 4log⁡x34\log_{x}3. Change of base formula: log⁡ab=log⁡(c)blog⁡(c)a\log_{a}b = \frac{\log(c)b}{\log(c)a} 4log⁡x3=4×(log⁡(3)log⁡(x))4\log_{x}3 = 4 \times \left(\frac{\log(3)}{\log(x)}\right)
  3. Simplify right side: Simplify the right side of the equation.\newlineWe have (1)/(log⁡93)(1)/(\log_{9}3). Using the change of base formula again, we get:\newline(1)/(log⁡93)=log⁡(3)/log⁡(9)(1)/(\log_{9}3) = \log(3) / \log(9)\newlineSince 99 is 323^2, log⁡(9)=log⁡(32)=2log⁡(3)\log(9) = \log(3^2) = 2\log(3), so:\newline(1)/(log⁡93)=log⁡(3)/(2log⁡(3))=1/2(1)/(\log_{9}3) = \log(3) / (2\log(3)) = 1/2
  4. Substitute simplified expressions: Substitute the simplified expressions back into the equation.\newlineNow we have log⁡3x−4×(log⁡(3)log⁡(x))=1+12\log_{3}x - 4 \times \left(\frac{\log(3)}{\log(x)}\right) = 1 + \frac{1}{2}
  5. Combine constants on right side: Combine the constants on the right side of the equation.\newline1+12=321 + \frac{1}{2} = \frac{3}{2}\newlineSo, the equation becomes log⁡3x−4⋅(log⁡(3)log⁡(x))=32\log_{3}x - 4 \cdot \left(\frac{\log(3)}{\log(x)}\right) = \frac{3}{2}
  6. Convert to natural logarithm: Convert log⁡3x\log_{3}x to the natural logarithm using the change of base formula.\newlinelog⁡3x=log⁡(x)log⁡(3)\log_{3}x = \frac{\log(x)}{\log(3)}\newlineThe equation now is (log⁡(x)log⁡(3))−4⋅(log⁡(3)log⁡(x))=32\left(\frac{\log(x)}{\log(3)}\right) - 4 \cdot \left(\frac{\log(3)}{\log(x)}\right) = \frac{3}{2}
  7. Multiply through to clear denominators: Multiply through by log⁡(3)log⁡(x)\log(3)\log(x) to clear the denominators.$log⁡(x)⋅log⁡(x)\$\log(x) \cdot \log(x) - 44 \cdot (\log(33) \cdot \log(33)) = \left(\frac{33}{22}\right) \cdot \log(33) \cdot \log(x)\)
  8. Simplify the equation: Simplify the equation.\newline(log⁡(x))2−4⋅(log⁡(3))2=32⋅log⁡(3)⋅log⁡(x)(\log(x))^2 - 4 \cdot (\log(3))^2 = \frac{3}{2} \cdot \log(3) \cdot \log(x)
  9. Rearrange into quadratic form: Rearrange the equation into a quadratic form.\newlineLet's set A=log⁡(x)A = \log(x) and B=log⁡(3)B = \log(3), then we have:\newlineA2−32⋅B⋅A−4⋅B2=0A^2 - \frac{3}{2} \cdot B \cdot A - 4 \cdot B^2 = 0
  10. Solve quadratic equation for A: Solve the quadratic equation for A.\newlineThis is a quadratic equation in the form of A2−C⋅A−D=0A^2 - C \cdot A - D = 0, where C=(32)⋅BC = \left(\frac{3}{2}\right) \cdot B and D=4⋅B2D = 4 \cdot B^2.\newlineWe can use the quadratic formula to solve for A: A=C±C2+4D2A = \frac{C \pm \sqrt{C^2 + 4D}}{2}
  11. Calculate discriminant: Calculate the discriminant of the quadratic equation.\newlineThe discriminant is C2+4D=(32⋅B)2+4⋅4⋅B2C^2 + 4D = \left(\frac{3}{2} \cdot B\right)^2 + 4 \cdot 4 \cdot B^2
  12. Substitute BB into discriminant: Substitute B=log⁡(3)B = \log(3) into the discriminant.\newlineDiscriminant = (32⋅log⁡(3))2+4⋅4⋅(log⁡(3))2\left(\frac{3}{2} \cdot \log(3)\right)^2 + 4 \cdot 4 \cdot (\log(3))^2
  13. Simplify the discriminant: Simplify the discriminant.\newlineDiscriminant = (94)⋅(log⁡(3))2+16⋅(log⁡(3))2(\frac{9}{4}) \cdot (\log(3))^2 + 16 \cdot (\log(3))^2\newlineDiscriminant = (94+16)⋅(log⁡(3))2(\frac{9}{4} + 16) \cdot (\log(3))^2\newlineDiscriminant = (94+644)⋅(log⁡(3))2(\frac{9}{4} + \frac{64}{4}) \cdot (\log(3))^2\newlineDiscriminant = (\frac{\(73\)}{\(4\)}) \cdot (\log(\(3))^22
  14. Solve for A using quadratic formula: Solve for A using the quadratic formula.\newlineA=(32)∗B±(734)∗B22A = \frac{\left(\frac{3}{2}\right) * B \pm \sqrt{\left(\frac{73}{4}\right) * B^2}}{2}\newlineA=(32)∗log⁡(3)±(734)∗(log⁡(3))22A = \frac{\left(\frac{3}{2}\right) * \log(3) \pm \sqrt{\left(\frac{73}{4}\right) * (\log(3))^2}}{2}
  15. Solve for x: Since A=log⁡(x)A = \log(x), we can solve for x.x=10Ax = 10^Ax=10(32⋅log⁡(3)±734⋅(log⁡(3))2)/2x = 10^{\left(\frac{3}{2} \cdot \log(3) \pm \sqrt{\frac{73}{4} \cdot (\log(3))^2}\right) / 2}
  16. Realize mistake in previous step: Realize there is a mistake in the previous step.\newlineThe base of the logarithm is 33, not 1010. Therefore, we should use 33 as the base for exponentiation, not 1010.\newlinex=3Ax = 3^A\newlinex=3(32⋅log⁡(3)±734⋅(log⁡(3))2)/2x = 3^{\left(\frac{3}{2} \cdot \log(3) \pm \sqrt{\frac{73}{4} \cdot (\log(3))^2}\right) / 2}

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