If log ₃2,log ₃(2ˣ - 5),log ₃( 2ˣ - 7/2 ) are in an arithmetic progression, then the value of x is equal to _____________.

Solution

Maths sir · 4 steps

Three logs in AP means the middle one doubled equals the sum of the outer two, so drop the logs.

1. AP condition. For a, b, c in AP, 2b = a + c. Use 2 log₃ m = log₃ m² and log₃ p + log₃ q = log₃ pq, then compare the insides.

   (2ˣ - 5)² = 2(2ˣ - 7/2) = 2 · 2ˣ - 7

2. Substitute. Put t = 2ˣ. Then t² - 10t + 25 = 2t - 7, so t² - 12t + 32 = 0, which factors as (t - 4)(t - 8) = 0. So t = 4 or t = 8.
3. Check the domain. A log needs a positive argument, so 2ˣ - 5 > 0. With t = 4, 2ˣ - 5 = -1, which is not allowed. With t = 8, 2ˣ - 5 = 3 and 2ˣ - 7/2 = 9/2, both positive.
4. So. 2ˣ = 8, which gives x = 3. The terms are log₃ 2, log₃ 3 and log₃ 9/2, and they really are in AP.

The common slip: 2. Keeping the root 2ˣ = 4 gives x = 2, but then log₃(2ˣ - 5) = log₃(-1), which is not defined.

The idea to keep. After you remove the logs, every root must go back into the original logs. Any root that makes a log argument zero or negative gets thrown out.

Answer from the NTA final answer key. Solution by PrepWiser.