Question

Log3^2X+ 5log9 X – 1,5=0​

Answer 1

log3²X+ 5log9 X – 1,5=0. ОДЗ: x>0

log3²X + 5/2 log3X - 3/2 = 0|×2

2log3²X + 5 log3X - 3 = 0

Нехай, Iog3X = t, тоді

2t² + 5t - 3 = 0

t1 = -6/2. t = -3

t2 = 1/2. t = 1/2

Iog3X = -3. Iog3X = 1/2

x1 = 1/27. x2 = √3

Відповідь: 1/27; √3

Answer 2

To solve the equation $\log_3^2x + 5\log_9x - 1.5 = 0$, we can use some properties of logarithms to simplify it.

1. Use the change of base formula to express $\log_9x$ in terms of $\log_3x$: $\log_9x = \frac{\log_3x}{\log_39}$

2. Now, rewrite the equation with this substitution: $\log_3^2x + 5\left(\frac{\log_3x}{\log_39}\right) - 1.5 = 0$

3. To get rid of the fraction, multiply both sides of the equation by $\log_39$: $\log_3^2x\log_39 + 5\log_3x - 1.5\log_39 = 0$

4. Let $y = \log_3x$. Now, the equation becomes: $(\log_3y)^2\log_39 + 5\log_3y - 1.5\log_39 = 0$

5. Replace $\log_3y$ with $y$: $(y^2)\log_39 + 5y - 1.5\log_39 = 0$

6. Now, solve this quadratic equation for $y$: $(y^2)\log_39 + 5y - 1.5\log_39 = 0$

To solve for $y$, you can use the quadratic formula:

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

Where $a = \log_39$, $b = 5$, and $c = -1.5\log_39$.

Plug these values into the quadratic formula:

$y = \frac{-5 \pm \sqrt{5^2 - 4(\log_39)(-1.5\log_39)}}{2(\log_39)}$

Solve for $y$ using this formula, and once you have $y$, you can then find $x$ by reversing the substitution:

$x = 3^y$

This will give you the solutions for the equation $\log_3^2x + 5\log_9x - 1.5 = 0$.