Sure, let's solve the system of equations step by step.

Given equation is log2x + log2y = log29 + 1

Step 1: We know that loga + logb = log(ab). So, we can rewrite the equation as log2(xy) = log29 + 1

Step 2: We also know that loga^n = nloga. So, we can rewrite the equation as log2(xy) = log2(2^9) + log2(2)

Step 3: Simplifying the right side of the equation, we get log2(xy) = log2(512*2) = log2(1024)

Step 4: If loga = logb, then a = b. So, we can equate the arguments of the logarithms to get xy = 1024

Step 5: The problem doesn't specify any other relationship between x and y, so there are multiple solutions. For example, if x = 2 and y = 512, then xy = 1024. Similarly, if x = 32 and y = 32, then xy = 1024.

So, the solution set (p, q) could be any pair of positive numbers such that their product is 1024.

Question

Sure, let's solve the system of equations step by step.

Given equation is log2x + log2y = log29 + 1

Step 1: We know that loga + logb = log(ab). So, we can rewrite the equation as log2(xy) = log29 + 1

Step 2: We also know that loga^n = nloga. So, we can rewrite the equation as log2(xy) = log2(2^9) + log2(2)

Step 3: Simplifying the right side of the equation, we get log2(xy) = log2(512*2) = log2(1024)

Step 4: If loga = logb, then a = b. So, we can equate the arguments of the logarithms to get xy = 1024

Step 5: The problem doesn't specify any other relationship between x and y, so there are multiple solutions. For example, if x = 2 and y = 512, then xy = 1024. Similarly, if x = 32 and y = 32, then xy = 1024.

So, the solution set (p, q) could be any pair of positive numbers such that their product is 1024.

Knowee AI · Accepted Answer

Sure, let's solve the system of equations step by step.

Given equation is log2x + log2y = log29 + 1

Step 1: We know that loga + logb = log(ab). So, we can rewrite the equation as log2(xy) = log29 + 1

Step 2: We also know that loga^n = nloga. So, we can rewrite the equation as log2(xy) = log2(2^9) + log2(2)

Step 3: Simplifying the right side of the equation, we get log2(xy) = log2(512*2) = log2(1024)

Step 4: If loga = logb, then a = b. So, we can equate the arguments of the logarithms to get xy = 1024

Step 5: The problem doesn't specify any other relationship between x and y, so there are multiple solutions. For example, if x = 2 and y = 512, then xy = 1024. Similarly, if x = 32 and y = 32, then xy = 1024.

So, the solution set (p, q) could be any pair of positive numbers such that their product is 1024.

The solution set of the system of equations log2x + log2y = log29 + 1, is (p, q) then

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