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#1

$Old$ October 6th, 2009, 12:33 PM

charmifs $charmifs is offline$

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$Default$ fractional indices

how do you solve

9^x = 27

thanks for any feedback

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#2

$Old$ October 6th, 2009, 12:45 PM

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how do you solve

9^x = 27

thanks for any feedback

I would suggest that you take a logarithm of both sides.

Remember that $\log_a (a^b) = b$

So proceed like so:

$\log_9 (9^x) = \log_9 (27)$

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#3

$Old$ October 6th, 2009, 12:49 PM

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Originally Posted by charmifs $View Post$

how do you solve

9^x = 27

thanks for any feedback

Another way is to use the powers of 3:

$(3^2)^x = 3^{2x} = 3^3$

If the bases are equal so too must the exponents:

$2x = 3$

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#4

$Old$ October 7th, 2009, 01:08 PM

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Thanks. That really helped.

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