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Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

\[3^{x -2}=27\]

OpenStudy (allank):

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

OpenStudy (anonymous):

What?!

OpenStudy (anonymous):

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

OpenStudy (allank):

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

OpenStudy (anonymous):

ohk

OpenStudy (allank):

@dpaInc 's method will also work.

OpenStudy (anonymous):

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

OpenStudy (allank):

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

OpenStudy (allank):

Are you okay with that step?

OpenStudy (anonymous):

i guess

OpenStudy (anonymous):

now i punch that into my calc?

OpenStudy (allank):

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

OpenStudy (allank):

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

OpenStudy (anonymous):

how does it keep changing like that?

OpenStudy (allank):

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

OpenStudy (allank):

Care to continue, @dpaInc ?

OpenStudy (anonymous):

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

OpenStudy (allank):

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

OpenStudy (anonymous):

im so confused

OpenStudy (anonymous):

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

OpenStudy (allank):

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

OpenStudy (anonymous):

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

OpenStudy (allank):

Glad to be of help :)

OpenStudy (anonymous):

can you help me with another one?

OpenStudy (allank):

Maybe later. Helpin out elsewhere. l8rz

OpenStudy (anonymous):

seriously? ugghhh!

OpenStudy (allank):

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

OpenStudy (anonymous):

lol i did and no help yet.

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