We can try taking the logarithm in base #7# of both sides:
#log_7(f(x))=log_7(7^(2x+7))#
we cancel the log in base #7# with the exponential of #7#;
#log_7(f(x))=cancel(log_7)(cancel(7)^(2x+7))#
and get:
#log_7(f(x))=2x+7#
isolate #x#:
#x=(log_7(f(x))-7)/2#
we change #x# with #F(x)# to write it as a normal function:
#F(x)=(log_7(x)-7)/2# and distinguish from the original one.
You can test your result using, say, #x=0.1#
so:
#f(0.1)=7^(2*0.1+7)=1,215,363#
if you use this value in the inverse you should get:
#F(1,215,363)=(log_7(1,215,363)-7)/2=0,1#...the original #x#!!!!
If we want we can even change the log into a natural log (#ln#, easier to evaluate using a pocket calculator) by using the Change of Base formula to get:
#F(x)=(color(red)(ln(x)/ln(7))-7)/2=(ln(x)-7ln(7))/(2ln(7))#
