CAT 2023 Slot 3 QA Question 1

First, the question says that three terms are in an arithmetic progression (AP). The terms are: 1/2, [log₃(2ˣ - 9)] / [log₃4], and [log₅(2ˣ + 17/2)] / [log₅4]. We find the common difference of this AP, and the options are given in terms of log base 4.

Hmm, okay. So, in an AP, the difference between consecutive terms is constant. That means the middle term minus the first term should be equal to the third term minus the middle term. So, if I denote the terms as a, b, c, then b - a = c - b, which implies 2b = a + c.

Let me write that down:

2 * [log₃(2ˣ - 9) / log₃4] = 1/2 + [log₅(2ˣ + 17/2) / log₅4]

Okay, that's the equation I need to solve. But before that, maybe I can simplify those logarithmic expressions. log_b a = (ln a) / (ln b), so [log_c d] / [log_c e] = log_e d. Is that right? Let me check.

Yes, because [log_c d] / [log_c e] = (ln d / ln c) / (ln e / ln c) = ln d / ln e = log_e d. So, that simplifies things.

So, applying that to the second term: [log₃(2ˣ - 9)] / [log₃4] = log_4 (2ˣ - 9). Similarly, the third term becomes log_4 (2ˣ + 17/2).

So now, the equation becomes:

2 * log_4 (2ˣ - 9) = 1/2 + log_4 (2ˣ + 17/2)

Hmm, that's better. Let me write that as:

2 log_4 (2ˣ - 9) - log_4 (2ˣ + 17/2) = 1/2

I can use logarithm properties to combine the terms on the left. Remember that 2 log_b a = log_b (a²), and log_b a - log_b c = log_b (a/c). So, applying that:

log_4 [(2ˣ - 9)² / (2ˣ + 17/2)] = 1/2

So, the left side is log base 4 of something equals 1/2. That means that something must be equal to 4^(1/2), which is 2. So,

(2ˣ - 9)² / (2ˣ + 17/2) = 2

Let me write that equation:

(2ˣ - 9)² = 2 * (2ˣ + 17/2)

Simplify the right side:

2 * (2ˣ + 17/2) = 2^(x+1) + 17

So, expanding the left side:

(2ˣ - 9)² = 2^(2x) - 18 * 2ˣ + 81

So, putting it all together:

2^(2x) - 18 * 2ˣ + 81 = 2^(x+1) + 17

Let me bring all terms to one side:

2^(2x) - 18 * 2ˣ - 2^(x+1) + 81 - 17 = 0

Simplify constants:

81 - 17 = 64

So,

2^(2x) - 18 2ˣ - 2 2ˣ + 64 = 0

Combine like terms:

-18 2ˣ - 2 2ˣ = -20 * 2ˣ

So,

2^(2x) - 20 * 2ˣ + 64 = 0

Hmm, this looks like a quadratic in terms of 2ˣ. Let me set y = 2ˣ. Then the equation becomes:

y² - 20y + 64 = 0

Let me solve this quadratic equation. The discriminant D = 400 - 256 = 144

So, y = [20 ± \] / 2 = [20 ± 12]/2

So, two solutions:

y = (20 + 12)/2 = 32/2 = 16

y = (20 - 12)/2 = 8/2 = 4

So, y = 16 or y = 4. Since y = 2ˣ, we have:

Now, I need to check if these x values satisfy the original logarithmic expressions. Because sometimes when we square or manipulate equations, we might introduce extraneous solutions.

Let's check x = 4:

First term: 1/2

Second term: log_4 (2⁴ - 9) = log_4 (16 - 9) = log_4 7

Third term: log_4 (2⁴ + 17/2) = log_4 (16 + 8.5) = log_4 24.5 = log_4 (49/2) = log_4 49 - log_4 2 = 2.5 - 0.5 = 2

Wait, Compute that again:

2⁴ = 16

2⁴ + 17/2 = 16 + 8.5 = 24.5 = 49/2

So, log_4 (49/2) = log_4 49 - log_4 2 = (log_4 7²) - 1 = 2 log_4 7 - 1

Hmm, that's not a nice number. Maybe I made a mistake.

Wait, 49 is 7², so log_4 49 = 2 log_4 7, and log_4 2 = 0.5 because 4^0.5 = 2.

So, log_4 (49/2) = 2 log_4 7 - 0.5

So, the terms would be:

1/2, log_4 7, 2 log_4 7 - 0.5

Now, let's check if they are in AP.

Compute the differences:

Second term - first term =