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anonymous · Answer

translate as :
solve for $x$ $$3^x=27$$

anonymous · Answer

are you sure you have to find the inverse of 3^x

anonymous · Answer

@Cosmichaotic

anonymous · Answer

no just solve it in your head

anonymous · Answer

$$\large 3^x=27$$what must $x$ be?

anonymous · Answer

f(x) = 3^x so to solve for f^-1(x) we must...

Replace f(x) with y, and switch the x and y

y = 3^x -> x = 3^y

Then solve for y.

anonymous · Answer

i can't wait to see this...

the_fizicx99 · Answer

xD

anonymous · Answer

Wait, am I doing this wrong?

anonymous · Answer

I was just solving for the inverse so we could plut in 27 for the new x?

anonymous · Answer

what you are going to do is call 
$$f^{-1}(*x)=\log_3(x)$$ which begs the question

anonymous · Answer

the answer is 3?

anonymous · Answer

I'm going to go do a chapter on logs and rules and stuff, I feel like I want to cry right now.

anonymous · Answer

once you have written 
$$\log_3(x)$$ and are asked to find $$\log_3(27)$$ you are gong to do it by solving 
$$3^x=27$$ which was the question to begin with

anonymous · Answer

yes, the answer is 3

anonymous · Answer

thanks guys

anonymous · Answer

but don't fret @Cosmichaotic  i was just pointing out that you cannot solve for the inverse of an exponential using algebra
saying 
$$3^x=27\iff \log_3(27)=x$$ is just saying the same thing back and forth

anonymous · Answer

like when a math teacher says 
solve $x^2=2$ and you are supposed to repeat 
$$x=\pm\sqrt2$$no information has passed
they say the same thing exactly

anonymous · Answer

Ah, I see - I have been working on the problem $$b ^{1.5} = 8 \rightarrow \log _{b}8 =1.5 $$ and trying to solve for b for 45 minutes. Lol.