How do you solve #4.5>=e^(0.031t)#?

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Answer 1 · 2016-11-01T12:15:48

#=> t<=ln(4.5)/0.031#

#=> t<=48.52# to 2 decimal places

As soon as you see the word solve it indicates you need to find at least 1 value.

Note that #e# is a predefined constant. So the only unknown is #t#

Also for all #ainRR: log_a(a)=1# and we can use this to 'get rid' of #e#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:#" "4.5>=e^(0.031t)#

Take logs of both sides

#ln(4.5)>=0.031t ln(e)#

But #ln(e)->log_e(e) =1# giving

#ln(4.5) >=0.031t#

#=> t<=ln(4.5)/0.031#

Tony B