How do you solve #\log 1.5-n\log 2<\log 5-3#?

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Answer 1 · 2017-09-10T23:51:50

#n > 8.228818690#

#color(blue)(9 .d.p.)#

Rearranging:

#- nlog2 < log 5 - log (1.5) - 3#

By law of logarithms #log(a) - log(b) = log(a/b)#

#-nlog2 < (log(5/1.5) - 3 )#

Divide by #log2# ( This will not affect the inequality sign, since logarithms greater than #1# are positive.)

#-n < (log(5/1.5) - 3 )/(log20#

Multiplying by #-1# and reversing inequality sign.

#n > (log(5/1.5) - 3 )/(log20#

Evaluate: #n > 8.228818690#

#color(blue)(9 .d.p.)#