Question

Log3(5x+3)+log3(1-2x)=2

Answer 1

Ответ:

ОДЗ: под логарифмическое выражение положительно, т.е.

\displaystyle \left \{ {{2x-5>0} \atop {2x-3>0}} \right.~~~\Rightarrow~~~\left \{ {{x>2.5} \atop {x>1.5}} \right.~~~\Rightarrow~~~ \boxed{x>2.5}

\log_3((2x-5)(2x-3))=\log_33\\ (2x-5)(2x-3)=3\\ 4x^2-16x+15=3\\ 4x^2-16x+12=0~~|:4\\ x^2-4x+3=0

По т. Виета:                 x_1=1 - не удовлетворяет ОДЗ

                                     x_2=3

Ответ: 3.

Answer 2

To solve the equation $\log_3(5x+3) + \log_3(1-2x) = 2$, you can use the properties of logarithms. First, we'll use the product rule of logarithms, which states that $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$. In this case, the product rule allows us to combine the two logarithms on the left side of the equation:

$\log_3((5x+3) \cdot (1-2x)) = 2$.

Now, we can rewrite the equation using the definition of logarithms:

$3^2 = (5x+3)(1-2x)$.

Simplify the equation:

$9 = (5x+3)(1-2x)$.

Expand the right side of the equation:

$9 = 5x - 10x^2 + 3 - 6x$.

Combine like terms:

$0 = -10x^2 - x + 12$.

Now, you have a quadratic equation. To solve for $x$, you can use the quadratic formula:

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

In this case, $a = -10$, $b = -1$, and (c = 12. Plug these values into the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-10)(12)}}{2(-10)}$.

Simplify:

$x = \frac{1 \pm \sqrt{1 + 480}}{-20}$.

$x = \frac{1 \pm \sqrt{481}}{-20}$.

So, the solutions for $x$ are:

$x = \frac{1 + \sqrt{481}}{-20}$ and $x = \frac{1 - \sqrt{481}}{-20}$.

These are the exact solutions for $x$. If you want decimal approximations, you can calculate them using a calculator.