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
\[3^{x -2}=27\]
Hint: try introducing logarithms to both sides of the equals sign.
What?!
can you write the 27 on the right hand side as a power of 3?
Do this: \[\log 3^{x-2} = \log 27\]
ohk
@dpaInc 's method will also work.
how do i do the ^x-2 on my calc?
That was the first step. It then becomes \[(x-2) \log 3 = \log 27 \]
Are you okay with that step?
i guess
now i punch that into my calc?
which then becomes \[(x-2) = \log27/\log3\]
So you can now work out log27/log3 using your calc
how does it keep changing like that?
These are logarithmic rules. If you aren't yet comfortable with them, it's better to use @dpaInc 's method
Care to continue, @dpaInc ?
well i did that log thing and i got 3 but wat is @dpaInc 's method?
So you got x-2=3 , then x=5 @dpaInc 's method involves writing the 27 as 3^(something). So, 27 = 3^?
im so confused
ohk i got that x-2=3 how does x=5?
You need to get x on one side of the equation, so you add 2 to both sides.
ohhh i got it! thanks @allank that was easy!
Glad to be of help :)
can you help me with another one?
Maybe later. Helpin out elsewhere. l8rz
seriously? ugghhh!
Take it easy. If you post another question in a new thread, I'm sure you'll get help. lol
lol i did and no help yet.
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