For what positive integer value will #2^x# first exceed #3x +2#?

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Answer 1 · 2018-01-16T22:46:42

#x = 4#

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the equation of the graph in red is #y = 2^x#
the equation of the graph in blue is #y = 3x+2#

their point of intersection is the value of #x# at which #2^x = 3x+2#.

for values of #x# higher than #3.717#, #2^x# has exceeded #3x+2# in terms of #y-#values.

on the graph, the #x-#value for the point of intersection is rounded to #3.717#.

the closest positive integer above this is #4.#

#4# is the first positive integer, for which #2^x > 3x+2#.
(#16 > 14#)