Question

Igx + lg(x + 1) = lg(5- 7x) – Ig2​

Answer 1

Ответ:

glx+log10(десятка снизу)(x+1)-2gl

Answer 2

To solve the equation $lg(x) + lg(x + 1) = lg(5 - 7x) - lg(2)$, let's first rewrite it using logarithmic properties:

$lg(x) + lg(x + 1) = lg\left(\frac{5 - 7x}{2}\right)$

Now, we can use logarithmic properties to combine the logarithms on the left side:

$lg(x(x + 1)) = lg\left(\frac{5 - 7x}{2}\right)$

Next, we can remove the logarithms by taking the base 10 exponent of both sides:

$x(x + 1) = \frac{5 - 7x}{2}$

Now, let's solve for $x$:

Expand and simplify the equation:

$x^2 + x = \frac{5 - 7x}{2}$

Multiply both sides by 2 to eliminate the fraction:

$2x^2 + 2x = 5 - 7x$

Rearrange to get a quadratic equation:

$2x^2 + 9x - 5 = 0$

Now, we can solve this quadratic equation using the quadratic formula:

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

where $a = 2$, $b = 9$, and $c = -5$.

$x = \frac{-9 \pm \sqrt{9^2 - 4 \times 2 \times -5}}{2 \times 2}$

$x = \frac{-9 \pm \sqrt{81 + 40}}{4}$

$x = \frac{-9 \pm \sqrt{121}}{4}$

$x = \frac{-9 \pm 11}{4}$

Now, we have two possible solutions for $x$:

$x_1 = \frac{-9 + 11}{4} = \frac{1}{2}$

$x_2 = \frac{-9 - 11}{4} = -5$

However, we need to check if these solutions are valid by ensuring that the logarithms in the original equation are defined for these values of $x$. Since logarithms are only defined for positive arguments, $x = -5$ is not a valid solution.

Therefore, the valid solution for the equation is $x = \frac{1}{2}$.