Plato is correct. If you want f of g, you just plug the g function into the x spot on the f function.
$\displaystyle f=number^x$
$\displaystyle g=blahx$
then $\displaystyle f\circ g(x)=number^{blahx}$
Does this make sense?
If I am incorrect on this please correct me!
Hello, Eraser147!
$\displaystyle \text{Given: }\begin{Bmatrix}f(x) &=& 7^{2+5x} \\ g(x) &=& \log_7(x) \end{Bmatrix} \quad \text{ Find }f\!\!\circ\! g.$
$\displaystyle f\!\!\circ\!g \;=\;f(g(x))$
. . . $\displaystyle =\;f(\log_7x)$
. . . $\displaystyle =\;7^{2+5\log_7(x)}$
. . . $\displaystyle =\;7^2\cdot 7^{5\log_7(x)} $
. . . $\displaystyle =\;49\cdot 7^{\log_7(x^5)} $
. . . $\displaystyle =\;49\cdot x^5$
. . . $\displaystyle =\;49x^5$