Question

Log 1/3 (27) + log 1/3 (1/9)

Answer 1

\(\huge \log_{\frac{1}{3}}27+\log_{\frac{1}{3}}\frac{1}{9}\)?

Answer 2

Yeah, What's the answer?

Answer 3

Then use the identity \(\large \log_BA+\log_BC=\log_BAC\)

Answer 4

-____-

Answer 5

I dont have a scientific calculator to do that, that's why I'm asking for the answer

Answer 6

Alyssa, OpenStudy is a place to learn, we will take the time with you to make sure you understand what you are asking for.

Answer 7

Following the identity I wrote above, you can simplify that as \(\large \log_{\frac{1}{3}}(27*\frac{1}{9})\).

Answer 8

Log 1/3 (3)=9?

Answer 9

i think its -1=Log 1/3 (3)

Answer 10

And \(\frac{1}{9}*27\) could be written as \(\frac{27}{9}\) which is 3

Answer 11

isn't \[\log _{\frac{1}{3}} 27 = (\frac{1}{3})^{-3}\]

and\[ \log _{\frac{1}{3}} 9 = (\frac{1}{3})^{-2}\]

Answer 12

So was I correct?

Answer 13

\(\large \log_\frac{1}{3}3\) would become the thing you simplify.

Answer 14

The answer isn't 9, let's see;

Answer 15

If we had \(\log_33\), it would be 1 right?

Answer 16

but the problem is, we have a fraction as a base, 3's reciprocal, therefore, add a minus sign: -1.

Answer 17

Okay

Answer 18

the problem is \[\log _{\frac{1}{3}} (27 \times \frac{1}{9}) = \log _{\frac{1}{3}} 3 = (\frac{1}{3})^{-1}\]

Answer 19

Log 1/3 (27) + log 1/3 (1/9)

Log 1/3 (27) = 3log 1/3 (3)
log 1/3 (1/9)= -2log 1/3 (3)

3log 1/3 (3) -2log 1/3 (3) = log 1/3 (3) = -1