Given that #ab^x<k#, for some constant k>0, show that #x>log(1600/k)/log2# where log means log to any valid base.(4 marks)? Given #a = 1600 # and #b = 1/2 #

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Answer 1 · 2018-06-11T16:12:04

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Use the log laws:

#color(red)(loga + logb = logab #

#color(red)(-loga = log(1/a) #

#color(red)(k log a = loga^k #

#1600(1/2)^x < k #

#=> log 1600(1/2)^x < log k #

#=> log1600 + log(1/2)^x < logk #

#=> log1600 + xlog(1/2) < logk #

#=> xlog(1/2) < logk - log1600#

#=> -xlog2 < logk - log1600 #

#=> xlog2 > log1600 - logk #

#=> xlog2 > log(1600/k) #

#=> x > ( log(1600/k) ) / log2 #