how do you solve 9^x = 27 thanks for any feedback
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Originally Posted by charmifs how do you solve 9^x = 27 thanks for any feedback I would suggest that you take a logarithm of both sides. Remember that $\displaystyle \log_a (a^b) = b $ So proceed like so: $\displaystyle \log_9 (9^x) = \log_9 (27)$
Originally Posted by charmifs how do you solve 9^x = 27 thanks for any feedback Another way is to use the powers of 3: $\displaystyle (3^2)^x = 3^{2x} = 3^3$ If the bases are equal so too must the exponents: $\displaystyle 2x = 3$
Thanks. That really helped.
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