Question

evaluate log base 9 of (1/27)
do u divide 9 into the fraction?

Answer 1

\[\log_9 \left(\frac{1}{27}\right) = x\] Now you're looking for a value of x that will turn 9 into 1/27, so \[\ 9^x=\frac{1}{27}\]

Answer 2

it would be a fraction?

Answer 3

Remember, if x is a number, then \(\ x^{-1} = \frac{1}{x}\)

Answer 4

so it would be -3?

Answer 5

\[\log_{9}(1/27)=-\log_{9}(27)=-\frac{\log_3(27)}{\log_{3}(9)}
\]

Answer 6

\[\ 9^{-3} = \frac{1}{729}\]

Answer 7

so it would be a fraction?

Answer 8

But look! I just realized something, haha.

9 is a factor of 729, is it not?

Answer 9

\(\bf \large {
log_{\color{red}{ a}}{\color{blue}{ b}}=y\iff {\color{red}{ a}}^y={\color{blue}{ b}}\qquad thus
\\ \quad \\
log_{\color{red}{ 9}}{\color{blue}{\frac{1}{27} }}=\square\implies {\color{red}{ 9}}^\square ={\color{blue}{ \frac{1}{27}}}\quad thus \quad \square =?
\\ \quad \\
\textit{bear in mind that } \cfrac{1}{a^{\color{red} n}}\implies a^{-{\color{red} n}}\quad thus
\\ \quad \\
9^\square =\cfrac{1}{27} \implies (3^2)^\square =\cfrac{1}{3^3}\implies 3^{2\square }=3^{-3}
}\)

Answer 10

it is so what if it is though, from what this guy is saying, it would be the base with a exponent

Answer 11

i did this in wolframalpha, and it came up with -3/2

Answer 12

I got it. I figured out an alternative way of explaining how @Zarkon derived the function.

Answer 13

ok

Answer 14

Using logarthmic rules, we've concluded \[9^x = \frac{1}{27}\]3 is a common factor between both of these, and we know in order to evaluate functions that are powers, we need to same base. rewrite 1/27 as 27^{-1} \[9^x = 27^{-1}\] Now set the base to 3. 9 3^2 and 27 is 3^3. \[ (3^2)^x = (3^3)^{-1}\]\[ 3^{2 \cdot x} = 3^{-1 \cdot 3}\]\[3^{2x} = 3^{-3}\]\[\color{red}3^{2x} = \color{red}3^{-3} \therefore \]\[2x = -3\] Can you solve for x now? :)

Answer 15

Sorry, typo. 9 = 3^2*

Answer 16

oh ok so thts how it equals -3/2

Answer 17

ok, thanks

Answer 18

Yeah, I don't quite remember how to solve using lug rules so I used what knowledge I had of log functions. took a little while to remember though!

Answer 19

well let me just say you're doing a fabulous job cause i just learned this today and i have a quiz on it tomorrow

Answer 20

so using this, log base 8 of 32 would be 4 right?

Answer 21

Practice these problems you're posting up on your own without looking at the solutions people have posted up here. That way it'll be ingrained into your head for tomorrow :)