We are given three equations and one inequality to solve: a. $(x+1)^{log(x+1)} = 100(x+1)$ b. $log_2 x = 1 + log_x 4$ c. $log_2 (x+3) + log_2 (x-3) = 2$ d. $log \frac{x^2-x}{x^2+1} < 0$

1. Problem Description

We are given three equations and one inequality to solve:

a. $(x+1)^{log(x+1)} = 100(x+1)$

b. $log_2 x = 1 + log_x 4$

c. $log_2 (x+3) + log_2 (x-3) = 2$

d. $log \frac{x^2-x}{x^2+1} < 0$

2. Solution Steps

a. $(x+1)^{log(x+1)} = 100(x+1)$

Take $log$ on both sides:

$log((x+1)^{log(x+1)}) = log(100(x+1))$

$log(x+1) * log(x+1) = log(100) + log(x+1)$

$(log(x+1))^2 = 2 + log(x+1)$

Let $y = log(x+1)$. Then the equation becomes:

$y^2 = 2 + y$

$y^2 - y - 2 = 0$

$(y-2)(y+1) = 0$

$y = 2$ or $y = -1$

Case 1: $log(x+1) = 2$

$x+1 = 10^2 = 100$

$x = 99$

Case 2: $log(x+1) = -1$

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

$x = \frac{1}{10} - 1 = -\frac{9}{10}$

b. $log_2 x = 1 + log_x 4$

$log_2 x = 1 + \frac{log_2 4}{log_2 x}$

$log_2 x = 1 + \frac{2}{log_2 x}$

Let $y = log_2 x$. Then the equation becomes:

$y = 1 + \frac{2}{y}$

$y^2 = y + 2$

$y^2 - y - 2 = 0$

$(y-2)(y+1) = 0$

$y = 2$ or $y = -1$

Case 1: $log_2 x = 2$

$x = 2^2 = 4$

Case 2: $log_2 x = -1$

$x = 2^{-1} = \frac{1}{2}$

c. $log_2 (x+3) + log_2 (x-3) = 2$

$log_2 ((x+3)(x-3)) = 2$

$log_2 (x^2 - 9) = 2$

$x^2 - 9 = 2^2 = 4$

$x^2 = 13$

$x = \pm\sqrt{13}$

Since $x > 3$, we have $x = \sqrt{13}$.

d. $log \frac{x^2-x}{x^2+1} < 0$

Since log is the base-10 logarithm, the expression is less than 0 when:

$\frac{x^2-x}{x^2+1} < 1$ and $\frac{x^2-x}{x^2+1} > 0$

$x^2 - x < x^2 + 1$

$-x < 1$

$x > -1$

$x^2 - x > 0$

$x(x-1) > 0$

So $x < 0$ or $x > 1$

Therefore, the solution is $(-1, 0) \cup (1, \infty)$

3. Final Answer

a. $x = 99$ or $x = -\frac{9}{10}$

b. $x = 4$ or $x = \frac{1}{2}$

c. $x = \sqrt{13}$

d. $x \in (-1, 0) \cup (1, \infty)$