Question

log[9] 27

Answer 1

is the base 9?

Answer 2

yes

Answer 3

I would divide up the log with the rules of logarithms.

Answer 4

log 27 / log 9

Answer 5

you could use this formula to simplify it:
\[\log_{b^{n}}(x) = \frac{1}{n}\log_b(x)\]

Answer 6

is says use the change-of-base property and a calculator is find a decimal appoximation

Answer 7

joe, what's that rule called? it's been a while since i did this stuff

Answer 8

so:
\[\log_9(27) = \log_{3^{2}}(27) = \frac{1}{2}\log_3(27) = \frac{1}{2}\log_3(3^{3}) = \frac{1}{2}*3 = \frac{3}{2}\]

Answer 9

i dont know what the rule is called to be honest, i just made it up based on what you said:
\[\log_9(27) = \frac{\log(27)}{\log(9)} = \frac{\log(27)}{2\log(3)} = \frac{1}{2}*\frac{\log(27)}{\log(3)} = \frac{1}{2}\log_3(27)\]

Answer 10

so its 3/2

Answer 11

yes :)

Answer 12

also like how 9^3/2 is the square root of 9^3 = 27

Answer 13

oh yeah, that works too, i wish i had seen it that way o.O

Answer 14

Ahh, cool! Thanks you guys, you rock my socks off!

Answer 15

it would become:
\[\log_9(27) = \log_9(9^{\frac{3}{2}}) = \frac{3}{2}\]
much shorter :)

Answer 16

yupp