Question

Lg^2 (x-1)-2lg(x-1)+1=0

Answer 1

ОДЗ: x>1

$log {}^{2} (x - 1) - 2 log(x - 1) + 1 = 0 \\ log(x - 1) = a \\ a {}^{2} - 2a + 1 = 0 \\ (a - 1) {}^{2} = 0 \\ a - 1 = 0 \\ a = 1$

сделаем обратную замену

$a = log(x - 1) \\ log(x - 1) = 1$

$x - 1 = 10 {}^{1} \\ x - 1 = 10 \\ x = 10 + 1 \\ x = 11$

$x = 11 \\ x > 1$

$x = 11 \\ otvet$

Answer 2

To solve the equation $\log^2(x - 1) - 2\log(x - 1) + 1 = 0$, we can use a substitution to simplify it. Let $y = \log(x - 1)$, then the equation becomes:

$y^2 - 2y + 1 = 0$.

Now, we have a quadratic equation in $y$. We can solve this quadratic equation for $y$:

$y^2 - 2y + 1 = 0$ factors as $(y - 1)^2 = 0$.

Taking the square root of both sides:

$y - 1 = 0$ or $y - 1 = 0$.

So, we have two solutions for $y$:

1. $y = 1$.
2. $y = 1$.

Now, we need to substitute back for $y$ in terms of $x$:

1. For $y = 1$, we have $\log(x - 1) = 1$. To solve for $x$, we can rewrite it in exponential form: $10^1 = x - 1$, which simplifies to $10 = x - 1$. Solving for $x$, we get $x = 11$.

2. For $y = 1$ (again), we have $\log(x - 1) = 1$, which gives the same solution as above: $x = 11$.

So, the equation $\log^2(x - 1) - 2\log(x - 1) + 1 = 0$ has one solution: $x = 11$.