Question

Log5(2х+3) =log5(x² +2х-1) .

Answer 1

Ответ: 2
_______________

Answer 2

To solve the equation $\log_5(2x+3) = \log_5(x^2 + 2x - 1)$, we can use the properties of logarithms. The key property we'll use here is that if two logarithms with the same base are equal, then their arguments must also be equal. In other words, if $\log_b(a) = \log_b(c)$, then $a = c$.

So, in your equation $\log_5(2x+3) = \log_5(x^2 + 2x - 1)$, we have:

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

Now, let's solve for $x$:

Subtract $2x$ from both sides:

$3 = x^2 - 1$

Add $1$ to both sides:

$4 = x^2$

Take the square root of both sides:

$x = \pm 2$

So, the solutions for the equation are $x = 2$ and $x = -2$. You can plug these values back into the original equation to verify that they satisfy the equation.