Question #cb13a

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Answer 1 · 2016-10-08T20:31:50

See the explanation section below.

#e^(0.1t) = 16#

Using #ln# for natural logarithm (logarithm base #e#).

#ln(e^(0.1t)) = ln(16)#

Now use #ln(e^r) = r# to get

#0.1t = ln(16)#

Multiply both sides by 10

#t = 10ln16#

We could rewrite this as #ln(16^10)# or as

#10ln16 = 10 ln(2^4) = 40 ln2#