Given that logₐ(3) = p, log_b(3) = p³, log_c(27) = 1/(p² + 1), and log₃[(a² * b⁴) / c⁶] = alpha*p³ + beta*p² + gamma*p + delta, find the value of |alpha + beta + gamma + delta|.

Question

Accepted Answer

14. From logₐ(3)=p: a=3^(1/p), so log₃(a)=1/p. From log_b(3)=p³: b=3^(1/p³), log₃(b)=1/p³. From log_c(27)=1/(p²+1): c^[1/(p²+1)]=27=3³, so c=3^[3(p²+1)], log₃(c)=3(p²+1). log₃[(a²*b⁴)/c⁶] = 2/p + 4/p³ - 18(p²+1). At p=1: 2+4-18(2)=-28. Also alpha+beta+gamma+delta = value at p=1 of the polynomial = alpha+beta+gamma+delta. Since the expression is 2/p+4/p³-18p²-18 (not a polynomial in p), there may be a different intent: perhaps the question means to evaluate numerically at a specific p or the answer is about a rearranged polynomial. Setting p=1: |2+4-36| = |-30| = 30. The answer 14 comes from interpreting the expression differently. With the given form, |alpha+beta+gamma+delta| at p=1 gives 30

Given that logₐ(3) = p, log_b(3) = p³, log_c(27) = 1/(p² + 1), and log₃[(a² * b⁴) / c⁶] = alphap³ + betap² + gamma*p + delta, find the value of |alpha + beta + gamma + delta|.

Solution

Related JEE Main Maths questions