Question

L og3(2x2+5 x−5)=l og3( 4+2 x )

Answer 1

Ответ:

После записи задачи будет так напиши следующее

$4 + 5x - 5 = 4 + 2x \\ 5x - 1 = 2 x + 4 \\ 5x - 2x = 4 + 1 \\ 3x = 5 \\ x = \frac{5}{3}$

Answer 2

To solve the equation $\log_3(2x^2 + 5x - 5) = \log_3(4 + 2x)$, you can use the property of logarithms that states that if $\log_a(b) = \log_a(c)$, then $b = c$.

So in your case, you can set the expressions inside the logarithms equal to each other:

$2x^2 + 5x - 5 = 4 + 2x$

Now, let's solve this equation for $x$:

First, move all terms to one side of the equation:

$2x^2 + 5x - 5 - 4 - 2x = 0$

Combine like terms:

$2x^2 + 3x - 9 = 0$

Now, you have a quadratic equation. You can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula is:

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

In your equation, $a = 2$, $b = 3$, and $c = -9$. Plug these values into the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-9)}}{2(2)}$

Now, calculate the discriminant ($b^2 - 4ac$):

$b^2 - 4ac = 3^2 - 4(2)(-9) = 9 + 72 = 81$

Now, substitute the discriminant back into the formula:

$x = \frac{-3 \pm \sqrt{81}}{4}$

Simplify the square root:

$x = \frac{-3 \pm 9}{4}$

Now, consider both the positive and negative square root solutions:

1. $x = \frac{-3 + 9}{4} = \frac{6}{4} = \frac{3}{2}$
2. $x = \frac{-3 - 9}{4} = \frac{-12}{4} = -3$

So, the solutions to the equation are $x = \frac{3}{2}$ and $x = -3$.