At x = 3 the exponential function h(x) begin to exceed the linear function f(x).Solution:Given data:[tex]f(x)=2x+25[/tex][tex]h(x)=5^x[/tex]To find at what value of x in h(x) begin to exceed the linear function f(x).Substitute x = 1 in f(x) and h(x).[tex]f(1)=2(1)+25=27[/tex][tex]h(1)=5^1=5[/tex]27 > 5f(x) is not exceed h(x), so proceed further.Substitute x = 2 in f(x) and h(x).[tex]f(2)=2(2)+25=29[/tex][tex]h(2)=5^2=25[/tex]29 > 25f(x) is not exceed h(x), so proceed further.Substitute x = 3 in f(x) and h(x).[tex]f(3)=2(3)+25=31[/tex][tex]h(3)=5^3=125[/tex]31 < 125f(x) is exceed h(x), the process stops here.Here the value of x is 3.Hence at x = 3 the exponential function h(x) begin to exceed the linear function f(x).