how do i solve this? i have to solve for x.
3^x^-2=27
x and -2 are exponents give me time to write it out

Question

anonymous · Answer

$$3^{x -2}=27$$

allank · Answer

Hint: try introducing logarithms to both sides of the equals sign.

anonymous · Answer

What?!

anonymous · Answer

can you write the 27 on the right hand side as a power of 3?

allank · Answer

Do this: 
$$\log 3^{x-2} = \log 27$$

anonymous · Answer

ohk

allank · Answer

@dpaInc 's method will also work.

anonymous · Answer

how do i do the ^x-2 on my calc?

allank · Answer

That was the first step. It then becomes 
$$(x-2) \log 3 = \log 27 $$

allank · Answer

Are you okay with that step?

anonymous · Answer

i guess

anonymous · Answer

now i punch that into my calc?

allank · Answer

which then becomes
$$(x-2) = \log27/\log3$$

allank · Answer

So you can now work out log27/log3 using your calc

anonymous · Answer

how does it keep changing like that?

allank · Answer

These are logarithmic rules. If you aren't yet comfortable with them, it's better to use @dpaInc 's method

allank · Answer

Care to continue, @dpaInc ?

anonymous · Answer

well i did that log thing and i got 3 but wat is @dpaInc 's method?

allank · Answer

So you got x-2=3 , then x=5
@dpaInc 's method involves writing the 27 as 3^(something). So, 27 = 3^?

anonymous · Answer

im so confused

anonymous · Answer

ohk i got that x-2=3 how does x=5?

allank · Answer

You need to get x on one side of the equation, so you add 2 to both sides.

anonymous · Answer

ohhh i got it! thanks @allank that was easy!

allank · Answer

Glad to be of help :)

anonymous · Answer

can you help me with another one?

allank · Answer

Maybe later. Helpin out elsewhere. l8rz

anonymous · Answer

seriously? ugghhh!

allank · Answer

Take it easy. If you post another question in a new thread, I'm sure you'll get help. lol

anonymous · Answer

lol i did and no help yet.