Find lowest value of X
Answer:
To find the lowest value of ( x ) for the given equation:
\frac{(2 \log_3 x) - 4}{\log_3 \left(\frac{x}{9}\right)} = \log_3 \left( x^7 \right) - \left( \frac{1}{\log_3 3} \right)^2 - 8
follow these steps:
-
Simplify the Right-Hand Side:
- We know that \log_3 \left( x^7 \right) = 7 \log_3 x
- \log_3 3 = 1 so \left( \frac{1}{\log_3 3} \right)^2 = 1
- Therefore the right-hand side can be simplified to:
7 \log_3 x - 1 - 8 = 7 \log_3 x - 9 -
Simplify the Left-Hand Side:
- Using the property of logarithms: \log_3 \left( \frac{x}{9} \right) = \log_3 x - \log_3 9
- Since \log_3 9 = 2 (because 9=3^2), we have \log_3 \left( \frac{x}{9} \right) = \log_3 x - 2.
- Thus, the left-hand side becomes:
\frac{2 \log_3 x - 4}{\log_3 x - 2} -
Equate the Simplified Expressions:
\frac{2 \log_3 x - 4}{\log_3 x - 2} = 7 \log_3 x - 9 -
Solve the Equation:
- Cross-multiplying to solve for (\log_3 x):
2 \log_3 x - 4 = (7 \log_3 x - 9)(\log_3 x - 2)- Expanding and simplifying:
2 \log_3 x - 4 = 7 (\log_3 x)^2 - 14 \log_3 x - 9 \log_3 x + 18- Combine like terms:
0 = 7 (\log_3 x)^2 - 14 \log_3 x - 11 \log_3 x + 22 + 2 \log_3 x - 40 = 7 (\log_3 x)^2 - 23 \log_3 x + 22 - 40 = 7 (\log_3 x)^2 - 23 \log_3 x + 18 -
Solve the Quadratic:
- Using the quadratic formula (ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}):
\log_3 x = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 7 \cdot 18}}{2 \cdot 7}\log_3 x = \frac{23 \pm \sqrt{529 - 504}}{14}\log_3 x = \frac{23 \pm \sqrt{25}}{14}\log_3 x = \frac{23 \pm 5}{14}\log_3 x = \frac{28}{14} \quad \text{or} \quad \log_3 x = \frac{18}{14}\log_3 x = 2 \quad \text{or} \quad \log_3 x = \frac{9}{7}
Therefore, the smallest value for (x) is:
\boxed{3^{9/7}}