Which point satisfies both #f(x)=2^x# and #g(x)=3^x#?

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Answer 1 · 2016-11-17T07:24:36

#(0, 1)#

If #f(x) = y = g(x)# then we have:

#2^x = 3^x#

Divide both sides by #2^x# to get:

#1 = 3^x/2^x = (3/2)^x#

Any non-zero number raised to the power #0# is equal to #1#. Hence #x=0# is a solution, resulting in:

#f(0) = g(0) = 1#

So the point #(0, 1)# satisfies #y = f(x)# and #y = g(x)#

Note also that since #3/2 > 1#, the function #(3/2)^x# is strictly monotonically increasing, so #x=0# is the only value for which #(3/2)^x = 1#