The image presents two problems. Problem 3 asks to solve the equation $\log_2(x+2) = \log_2(x-1) + 1$. Problem 4 asks to solve the equation $\log x = 3\log 1.2 + 2\log(\frac{1}{3}\sqrt{10}) - \log 0.96$.

1. Problem Description

The image presents two problems.

Problem 3 asks to solve the equation $\log_2(x+2) = \log_2(x-1) + 1$.

Problem 4 asks to solve the equation $\log x = 3\log 1.2 + 2\log(\frac{1}{3}\sqrt{10}) - \log 0.96$.

2. Solution Steps

Problem 3: $\log_2(x+2) = \log_2(x-1) + 1$

Step 1: Rewrite 1 as a logarithm with base

2. $1 = \log_2 2$

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

Step 2: Use the logarithm product rule $\log_a b + \log_a c = \log_a (bc)$.

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

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

Step 3: Since the logarithms are equal and have the same base, we can equate the arguments.

$x+2 = 2x-2$

Step 4: Solve for $x$.

$2x-x = 2+2$

$x = 4$

Step 5: Check the solution in the original equation.

$\log_2(4+2) = \log_2(4-1) + 1$

$\log_2(6) = \log_2(3) + 1$

$\log_2(6) = \log_2(3) + \log_2(2)$

$\log_2(6) = \log_2(3\times 2)$

$\log_2(6) = \log_2(6)$

The solution is valid.

Problem 4: $\log x = 3\log 1.2 + 2\log(\frac{1}{3}\sqrt{10}) - \log 0.96$

Step 1: Use the logarithm power rule $a \log b = \log b^a$.

$\log x = \log (1.2)^3 + \log (\frac{1}{3}\sqrt{10})^2 - \log 0.96$

$\log x = \log (1.728) + \log (\frac{1}{9} \times 10) - \log 0.96$

$\log x = \log (1.728) + \log (\frac{10}{9}) - \log 0.96$

Step 2: Use the logarithm product rule $\log a + \log b = \log (ab)$.

$\log x = \log (1.728 \times \frac{10}{9}) - \log 0.96$

$\log x = \log (1.92) - \log 0.96$

Step 3: Use the logarithm quotient rule $\log a - \log b = \log (\frac{a}{b})$.

$\log x = \log (\frac{1.92}{0.96})$

$\log x = \log 2$

Step 4: Since the logarithms are equal, we can equate the arguments.

$x = 2$

3. Final Answer

For problem 3, the solution is $x=4$.

For problem 4, the solution is $x=2$.