The problem asks us to solve three logarithmic equations for $x$. (a) $\log_{10}(x^2 + 13x) = 1 + \log_{10}(1+x)$ (b) $1 + \log_2(x^2 - 4x - 16) = \log_2(x^2 - 3x + 4)$ (c) $2(\log_2 x - 1) = \log_2 y$, where $y = x-1$

1. Problem Description

The problem asks us to solve three logarithmic equations for $x$.

(a) $\log_{10}(x^2 + 13x) = 1 + \log_{10}(1+x)$

(b) $1 + \log_2(x^2 - 4x - 16) = \log_2(x^2 - 3x + 4)$

(c) $2(\log_2 x - 1) = \log_2 y$, where $y = x-1$

2. Solution Steps

(a) $\log_{10}(x^2 + 13x) = 1 + \log_{10}(1+x)$

$\log_{10}(x^2 + 13x) = \log_{10} 10 + \log_{10}(1+x)$

$\log_{10}(x^2 + 13x) = \log_{10}(10(1+x))$

$x^2 + 13x = 10(1+x)$

$x^2 + 13x = 10 + 10x$

$x^2 + 3x - 10 = 0$

$(x+5)(x-2) = 0$

$x = -5$ or $x = 2$

Since the logarithm is only defined for positive arguments, we need to check the solutions.

If $x = -5$, then $1+x = -4$, which is not allowed since $\log_{10}(1+x)$ would not be defined.

If $x = 2$, then $x^2 + 13x = 4 + 26 = 30 > 0$ and $1+x = 3 > 0$. So $x = 2$ is a valid solution.

(b) $1 + \log_2(x^2 - 4x - 16) = \log_2(x^2 - 3x + 4)$

$\log_2 2 + \log_2(x^2 - 4x - 16) = \log_2(x^2 - 3x + 4)$

$\log_2(2(x^2 - 4x - 16)) = \log_2(x^2 - 3x + 4)$

$2(x^2 - 4x - 16) = x^2 - 3x + 4$

$2x^2 - 8x - 32 = x^2 - 3x + 4$

$x^2 - 5x - 36 = 0$

$(x-9)(x+4) = 0$

$x = 9$ or $x = -4$

If $x = 9$, then $x^2 - 4x - 16 = 81 - 36 - 16 = 29 > 0$ and $x^2 - 3x + 4 = 81 - 27 + 4 = 58 > 0$. So $x = 9$ is a valid solution.

If $x = -4$, then $x^2 - 4x - 16 = 16 + 16 - 16 = 16 > 0$ and $x^2 - 3x + 4 = 16 + 12 + 4 = 32 > 0$. So $x = -4$ is a valid solution.

(c) $2(\log_2 x - 1) = \log_2 y$, where $y = x-1$

$2\log_2 x - 2 = \log_2 (x-1)$

$\log_2 x^2 - \log_2 4 = \log_2 (x-1)$

$\log_2 \frac{x^2}{4} = \log_2 (x-1)$

$\frac{x^2}{4} = x-1$

$x^2 = 4x - 4$

$x^2 - 4x + 4 = 0$

$(x-2)^2 = 0$

$x = 2$

If $x = 2$, then $x > 0$ and $y = x - 1 = 2 - 1 = 1 > 0$. So $x = 2$ is a valid solution.

3. Final Answer

(a) $x = 2$

(b) $x = 9, x = -4$

(c) $x = 2$