Question

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

Answer 1

Решение задания на фото

Answer 2

To solve the equation $\log_3(2x - 3) + \log_3(x + 1) = 1$, you can use logarithm properties to combine the two logarithms on the left side and then solve for $x$. The properties you'll need are:

1. $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$
2. $\log_a(b) = 1$ is equivalent to $b = a$

Applying these properties to your equation:

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

Use the first property:

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

Now, rewrite the equation using the second property:

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

Simplify:

$(2x - 3)(x + 1) = 3$

Expand the left side:

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

Combine like terms:

$2x^2 - x - 6 = 3$

Subtract 3 from both sides:

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

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:

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

In the equation $2x^2 - x - 9 = 0$, you have $a = 2$, $b = -1$, and $c = -9$. Plug these values into the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-9)}}{2(2)}$

Simplify:

$x = \frac{1 \pm \sqrt{1 + 72}}{4}$

$x = \frac{1 \pm \sqrt{73}}{4}$

So, the solutions for $x$ are:

$x = \frac{1 + \sqrt{73}}{4}$ and $x = \frac{1 - \sqrt{73}}{4}$