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))
