if g(x)=log base7 X, then g^-1(x)=?

Question

anonymous · Answer

g^-1(x)=7^x

zion · Answer

how?

anonymous · Answer

nah, its $$\log_{x} 7$$

anonymous · Answer

working
y=logbase7 X
7^y=x
f^-1(x)=7^y

anonymous · Answer

youre asking about $$1 \div g (x) $$ right?

zion · Answer

hashir got the correct answer.

anonymous · Answer

yeah i am always right!!!!!!

zion · Answer

hahaha. stay humble dude! hahahahahaha!

anonymous · Answer

hahahahahah jk
fb id plz :D

zion · Answer

www.facebook.com/paulguarnes2 ... you cant add me on my main account coz its already full.

anonymous · Answer

ok

anonymous · Answer

anyway, take a look at this:: http://www.purplemath.com/modules/logrules.htm

anonymous · Answer

it is inverse boy !!!!!!!!!

anonymous · Answer

i think u need to learn functions

anonymous · Answer

+ log too

anonymous · Answer

right. got lots to learn. about understanding the coded lingo here. :P

zion · Answer

The questions that im asking are from my reviewer, and they got the correct answer, but the solutions are not provided. Hashir got the correct answer, all i wanna learn is how to get the answer.

anonymous · Answer

Yeah I got the Same

anonymous · Answer

Ok let me re-type the whole thing

anonymous · Answer

$$g(x) = log_7 x$$
Now Let g(x) = y
$$y = log_7x$$

anonymous · Answer

Hence to get Inverse we need to turn x as a function of y

anonymous · Answer

$$x = 7^y$$From$$\large{log_{base}x =y => x = {base}^y}$$

zion · Answer

the answer is 7^x

anonymous · Answer

Now we have the Setup that makes 

$$g^{-1}(x) = 7^x$$is the Inverse

anonymous · Answer

yeah you got it : )