# Formula for f of g

• October 9th 2012, 09:51 AM
Eraser147
Formula for f of g
Attachment 25132Click the image to enlarge please.
• October 9th 2012, 10:09 AM
Plato
Re: Formula for f of g
Quote:

Originally Posted by Eraser147
Attachment 25132Click the image to enlarge please.

$f\circ g(x)=7^{2+5\log_7(x)}$

BUT that can be greatly simplified.
• October 9th 2012, 10:31 AM
Eraser147
Re: Formula for f of g
Yes, but going from there, I don't know how to find the x. It's just too confusing from there. I don't know which exponent I can factor out.
• October 9th 2012, 10:34 AM
alane1994
Re: Formula for f of g
Plato is correct. If you want f of g, you just plug the g function into the x spot on the f function.
$f=number^x$
$g=blahx$
then $f\circ g(x)=number^{blahx}$
Does this make sense?
If I am incorrect on this please correct me!:)
• October 9th 2012, 10:40 AM
Plato
Re: Formula for f of g
Quote:

Originally Posted by Eraser147
Yes, but going from there, I don't know how to find the x. It's just too confusing from there. I don't know which exponent I can factor out.

Well I did say "that can be greatly simplified."

$f\circ g(x)=7^{2+5\log_7(x)}=49x^5$
• October 9th 2012, 10:54 AM
Soroban
Re: Formula for f of g
Hello, Eraser147!

Quote:

$\text{Given: }\begin{Bmatrix}f(x) &=& 7^{2+5x} \\ g(x) &=& \log_7(x) \end{Bmatrix} \quad \text{ Find }f\!\!\circ\! g.$

$f\!\!\circ\!g \;=\;f(g(x))$

. . . $=\;f(\log_7x)$

. . . $=\;7^{2+5\log_7(x)}$

. . . $=\;7^2\cdot 7^{5\log_7(x)}$

. . . $=\;49\cdot 7^{\log_7(x^5)}$

. . . $=\;49\cdot x^5$

. . . $=\;49x^5$
• October 9th 2012, 11:16 AM
Eraser147
Re: Formula for f of g
^ Thankyou that made more sense. I had to separate the 7s.
• October 9th 2012, 11:43 AM
Eraser147
Re: Formula for f of g
Just one more question, can someone explain why when 7 is extended to the power of log 7 and raised to the x^5, why does it cancel and equate to x^5?
• October 9th 2012, 12:00 PM
Plato
Re: Formula for f of g
Quote:

Originally Posted by Eraser147
Just one more question, can someone explain why when 7 is extended to the power of log 7 and raised to the x^5, why does it cancel and equate to x^5?

$b^{\log_b(a)}=a~\&~b^{n\log_b(a)}=b^{\log_b(a^n)}= a^n$
• October 9th 2012, 12:25 PM
Eraser147
Re: Formula for f of g
Thanks
• October 9th 2012, 12:30 PM
HallsofIvy
Re: Formula for f of g
That is, pretty much, the definition of " $log_a(x)$"- it is the inverse function to $a^x$: $a^{log_a(x)}= log_a(a^x)= x$.