The problem consists of two parts. Part 1: Solve the equation $5(1 + 4e^{0.1x-1}) = 25$ for $x$. Part 2: Solve the equation $log_2(x+6) + log_2(x) = 4$ for $x$.

1. Problem Description

The problem consists of two parts.

Part 1: Solve the equation $5(1 + 4e^{0.1x-1}) = 25$ for $x$.

Part 2: Solve the equation $log_2(x+6) + log_2(x) = 4$ for $x$.

2. Solution Steps

Part 1:

We want to solve $5(1 + 4e^{0.1x-1}) = 25$.

Divide both sides by 5:

$1 + 4e^{0.1x-1} = 5$

Subtract 1 from both sides:

$4e^{0.1x-1} = 4$

Divide both sides by 4:

$e^{0.1x-1} = 1$

Take the natural logarithm of both sides:

$ln(e^{0.1x-1}) = ln(1)$

$0.1x - 1 = 0$

$0.1x = 1$

$x = \frac{1}{0.1}$

$x = 10$

Part 2:

We want to solve $log_2(x+6) + log_2(x) = 4$.

Using the logarithm property $log_a(b) + log_a(c) = log_a(bc)$, we have:

$log_2((x+6)x) = 4$

$log_2(x^2 + 6x) = 4$

Exponentiate both sides with base 2:

$2^{log_2(x^2+6x)} = 2^4$

$x^2 + 6x = 16$

$x^2 + 6x - 16 = 0$

Factor the quadratic equation:

$(x+8)(x-2) = 0$

So, $x = -8$ or $x = 2$.

Since we have $log_2(x)$ in the original equation, $x$ must be positive. Therefore, $x = -8$ is not a valid solution.

So we have $x=2$.

We check $x=2$:

$log_2(2+6) + log_2(2) = log_2(8) + log_2(2) = 3 + 1 = 4$.

3. Final Answer

Part 1: $x = 10$

Part 2: $x = 2$