The problem has three questions. Question 1: Given the equation $3^{a-2} = 5$, find the value of $a$ that satisfies the equation from the options. Question 2: Find the solution of the radical equation $\sqrt{2x-3} - 5 = -2$ from the options. Question 3: Two consecutive odd positive integers are such that the sum of three times the smaller integer and twice the greater integer is 59. Find the greater integer from the options.

1. Problem Description

The problem has three questions.

Question 1: Given the equation $3^{a-2} = 5$, find the value of $a$ that satisfies the equation from the options.

Question 2: Find the solution of the radical equation $\sqrt{2x-3} - 5 = -2$ from the options.

Question 3: Two consecutive odd positive integers are such that the sum of three times the smaller integer and twice the greater integer is

5

9. Find the greater integer from the options.

2. Solution Steps

Question 1: $3^{a-2} = 5$

Take the logarithm of both sides (base 10 or natural logarithm):

$(a-2) \log(3) = \log(5)$

$a-2 = \frac{\log(5)}{\log(3)}$

$a = \frac{\log(5)}{\log(3)} + 2$

$a \approx \frac{0.699}{0.477} + 2 \approx 1.465 + 2 \approx 3.465$

The closest option is $3.45$.

Question 2: $\sqrt{2x-3} - 5 = -2$

$\sqrt{2x-3} = 3$

Square both sides:

$2x-3 = 9$

$2x = 12$

$x = 6$

The solution is $x=6$.

Question 3:

Let $x$ be the smaller odd positive integer. Then $x+2$ is the greater odd positive integer.

The sum of three times the smaller integer and twice the greater integer is

5

9. $3x + 2(x+2) = 59$

$3x + 2x + 4 = 59$

$5x = 55$

$x = 11$

The smaller integer is 11, and the greater integer is $x+2 = 11+2 = 13$.

The greater integer is

1

3.

3. Final Answer

1. A. 3.45

2. B. 6

3. D. 13