1. The sequence log2 32, log2 y, log2 128, ... forms an arithmetic sequence. What is the value of y?

2. If log a^2 b^3 = x and log (a/b) = y, what are the values of log a and log b?

128/32 = r^2

r=2, so y=32*2=64
Or, evaluate the logs. The sequence is thus
5,log2(y),7
so, log2(y) = 6, and y=64

log a^2b^3 = 2loga + 3logb
log a/b = loga - logb

2loga + 3logb = x
loga - logb = y

2loga + 3logb = x
2loga - 2logb = 2y
5logb = x-2y
logb = (x-2y)/5

solve for loga in like wise.

To find the value of y in the sequence log2 32, log2 y, log2 128, we need to determine the common difference of the arithmetic sequence.

Recall that in an arithmetic sequence, each term is obtained by adding a constant difference to the previous term.

In this case, we can find the common difference by subtracting any two consecutive terms. Let's subtract log2 32 from log2 y:

log2 y - log2 32 = log2 (y/32)

Similarly, let's subtract log2 128 from log2 y:

log2 y - log2 128 = log2 (y/128)

Since we know that these two expressions should be equal, we set them equal to each other:

log2 (y/32) = log2 (y/128)

Now, we can solve this equation:

y/32 = y/128

To simplify the equation, we can multiply both sides by the least common multiple of the denominators, which is 128:

128(y/32) = 128(y/128)

4y = y

Dividing both sides by y:

4 = 1

This equation is contradictory, so there is no valid value of y that satisfies the given arithmetic sequence.

Therefore, the value of y is undefined in this case.

Moving on to the second question:

To determine the values of log a and log b when given log a^2 b^3 = x and log (a/b) = y, we can use the properties of logarithms and algebraic manipulation.

Let's start with the first equation:

log a^2 b^3 = x

Using the property of logarithms that states log a^x = x log a, we can rewrite the equation as:

2 log a + 3 log b = x

Now, let's move on to the second equation:

log (a/b) = y

Using the property of logarithms that states log (a/b) = log a - log b, we can rewrite the equation as:

log a - log b = y

We now have a system of equations:

2 log a + 3 log b = x ---(1)
log a - log b = y ---(2)

To find the values of log a and log b, we can solve this system of equations simultaneously.

First, let's eliminate log b by multiplying equation (2) by 3:

3(log a - log b) = 3y
3log a - 3log b = 3y

Now, let's add this equation to equation (1):

2 log a + 3 log b + 3 log a - 3 log b = x + 3y

Simplifying the equation, we get:

5 log a = x + 3y

Now, divide both sides of the equation by 5:

log a = (x + 3y)/5

Next, substitute this expression for log a back into equation (2):

(x + 3y)/5 - log b = y

Multiply both sides by 5:

x + 3y - 5 log b = 5y

Rearrange the equation:

5 log b = x - 2y

Divide both sides by 5:

log b = (x - 2y)/5

Therefore, the values of log a and log b are (x + 3y)/5 and (x - 2y)/5, respectively.