If log 2 , log (x power 2 - 1) and log (x power 2 + 3 ) are in A.P. Then find x?

Given:- log2, log[(x^2)-1] and log[(x^2)+3] are in AP. To find: Values of x satisfying the above relation. Solution:- Step 1 If a,b,c are in AP then b - a = c - b. Step 2 for ease of solving let t=x^2 So, log(t-1) - log2 = log(t+3) - log(t-1) ..........(1) Step 3 log a - log b = log(a/b) applying this property to equation (1) log [(t -1)/2] = log [(t+3)/(t-1)]. Step 4 Taking anti-log on both sides, (t-1)/2 = (t+3)/(t-1), Step 5 Cross multiplying (t^2) - 2t + 1 = 2t + 6 taking all terms towards LHS t^2 - 4t -5 = 0 Step 6 Finding roots of this equation t^2 + (-5+1)t + [(-5)*1] = 0 (t-5)*(t+1) = 0 So, t = 5 and t = -1. Step 7 As we assumed before t = x^2 Case 1 t = 5 so x^2 = 5 therefore, x = +sqrt(5) or -sqrt(5) Case 2 t=-1 so x^2 = -1 therefore, x = +i or -i. So final values of x satisfying the above relation are x = +sqrt(5), -sqrt(5), +i, -i. NOTE:- Here sqrt() denotes the square root operation. 'i' denotes sqrt(-1), which is an imaginary number. read less