Rewrite the logarit… - QuestionCove
OpenStudy (anonymous):

Rewrite the logarithm in terms of variables given Log_9 (12) =a Log_9 (5) =b Log_9 (8)=c Rewrite: Log_9 (1/192)

2 years ago
Parth (parthkohli):

$192 = 12 \times 8 \times 8^{1/3}$

2 years ago
Parth (parthkohli):

$\log_9 \left(1/192\right)=-\log_9 (192) = -\log_9(12\times 8 \times 8^{1/3}) = \cdots$

2 years ago
OpenStudy (anonymous):

-A-B-(b/3)?

2 years ago
Parth (parthkohli):

Yup :)

2 years ago
OpenStudy (anonymous):

Oh so we do not include log _9 (1)?

2 years ago
Parth (parthkohli):

$\log(1/k) \\= \log(1) - \log(k)\\ = 0 - \log(k) \\= -\log(k)$

2 years ago
Parth (parthkohli):

or$\log(1/k) \\ = \log(k^{-1}) \\ = - 1\cdot \log(k) \\ = -\log(k)$

2 years ago
OpenStudy (anonymous):

Hogg. And also, how did you know to use 8^(1/3)?

2 years ago
OpenStudy (fibonaccichick666):

Experience, do enough of these and you see the tricks

2 years ago
OpenStudy (anonymous):

My brain wouldn't have thought about that. So would there be a way to figure it out

2 years ago
OpenStudy (anonymous):

Ok.

2 years ago
Parth (parthkohli):

I tried to express 192 as a product of powers of 12, 8 and 5. There can't be a 5 in there, so that is ruled out.$12 \times 8 = 96$I just need to multiply that by 2. I don't have 2 explicitly, but I can write that as the cube root of 8.

2 years ago
OpenStudy (anonymous):

Thank you. I couldn't figure out what to do with the last 2 XD

2 years ago