In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.
For exponential dependence we must have functions of form f(x) = a.(k power x) : a,k are constants and f(0)=a Let us assume: C = cycles for failure ; L = Load; a, k = constants according to question C is an exponential function of L Hence we can say, 1) C = a x (k^L) ---- > in case C is increasing exponentially 2) C = a / (k^L) ---- > in case C is decreasing exponentially we take 2nd equation and apply log to both side we get: log C + L x (log k) = log a ...... by logarithmic property given (C=100 for L=80) and (C=10000 for L=40) apply these to values to above equation, this gives us 2 equations and 2 variables: log a = log 100 + (80 x log k)...............(1) log a = log 10000 + (40 x log k)...........(2) we have 2 variables and 2 equations hence we find log a = 6 log k= 1/20 now find, [ L = (log a - log C) / (log k) ] for C=5000 cycles => L = (6 - log 5000)*20 = 46.0206 ~ 46.02 => [ANS]
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