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OpenstudyUser
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I thought you were soo smart, that you would know them by now. LOL jk jk.

saifoo.khan
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Lol. I'm not a calculator. I can get stuck too! :P
@OpenstudyUser

OpenstudyUser
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Try distributing.

saifoo.khan
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\[ \Large 9^{2x} + 2(9^{x+1}) = 40 \]
Correction^

saifoo.khan
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Just tell me the next and step after that please.

hartnn
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write 9^(x+1) as 9^x.9
then put y=9^x.
solve the quadratic

saifoo.khan
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DANG!!! KILL ME!!!!
My teacher told me that. _
Thanks for recalling @hartnn

saifoo.khan
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Just guessing, can't we do it like this?
\[9^{2x} + 2(9^x \times 9) = 40\]
Ans so on?

saifoo.khan
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Next? @micahwood50

hartnn
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didn't i tell u to do just that?

saifoo.khan
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Yes, but you said me to write 9^x = y
Got that.
But can't we solve it by another way?
Something like:
\[9^{2x} + 2(9^x)\times 18=40\]

hartnn
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nopes, i don't think so....no formula for log(a+b) or log (ab)

saifoo.khan
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I see. THanks man.

saifoo.khan
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Did you missed a +sign in your first step?
@micahwood50

saifoo.khan
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@OpenstudyUser kept on distributing. :D

OpenstudyUser
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I atleast tried _

micahwood50
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Just do what hartnn told you to do.

saifoo.khan
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I want to learn your method too! @micahwood50

micahwood50
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My "method" is incorrect. I just though lnab = ln (a + b), but it's actually lna + lnb so forgot it.

saifoo.khan
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Oh. No problem. Thanks for trying though. :)

saifoo.khan
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I solved it @hartnn .
got:
x = 0.315

hartnn
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what roots of quadratic did u get?

saifoo.khan
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2, 20

hartnn
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yup.
can't be 20
so 9^x=2
x=0.315
u are correct!

saifoo.khan
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Yep. Thanks.