Doing a logarithm exercise online. Multiple-choice, but I got an answer that isn't included? Extra conversion?

Question

I'm doing a logarithm excercise online, however I got a different answer to the ones provided (multiple-choice). Do I have to do extra conversions to my answer to reach one of the options?

This is the problem:

$$ \log _b\left(a^{-3}\right)+\left(\log _a\left(b^{-\frac{1}{6}}\right)\right)^{-1} $$

It's multiple choice question with the following options:

1. $ 3\log _a\left(b\right)$
2. $ 2\log _b\left(a\right)$
3. $ 3\log _b\left(a\right)$
4. $ 6\log _a\left(b\right)$

The answer I got was:

$$ -\frac{9}{\log \:_a\left(b\right)} $$

How I got it:

1. $ =-3\log _b\left(a\right)+\log _a\left(b^{-\frac{1}{6}}\right)^{-1} $
2. $ =-3\log _b\left(a\right)-\frac{1}{\frac{1}{6}\log _a\left(b\right)} $
3. Converted to fraction: $ =-\frac{1}{\frac{1}{6}\log _a\left(b\right)}-\frac{3\log _b\left(a\right)}{1} $
4. Adjusted based on the LCD: $ =-\frac{1}{\frac{1}{6}\log _a\left(b\right)}-\frac{3\log _b\left(a\right)\frac{1}{6}\log _a\left(b\right)}{\frac{1}{6}\log _a\left(b\right)} $
5. Combined: $ =\frac{-1-3\cdot \frac{1}{6}\log _b\left(a\right)\log _a\left(b\right)}{\frac{1}{6}\log _a\left(b\right)} $
6. Top: $ 3\cdot \frac{1}{6}\log _b\left(a\right)\log _a\left(b\right) $
7. $ =\frac{-1-\frac{1}{2}}{\frac{1}{6}\log _a\left(b\right)} $
8. Multiplied bottom: $ \frac{1}{6}\log _a\left(b\right)\::\quad \frac{\log _a\left(b\right)}{6} $
9. $ =\frac{-1-\frac{1}{2}}{\frac{\log _a\left(b\right)}{6}} $
10. $ =\frac{-\frac{3}{2}}{\frac{\log _a\left(b\right)}{6}} $
11. $ =-\frac{\frac{3}{2}}{\frac{\log _a\left(b\right)}{6}} $
12. $ =-\frac{3\cdot \:6}{2\log _a\left(b\right)} $
13. $ =-\frac{18}{2\log _a\left(b\right)} $
14. $ =-\frac{9}{\log _a\left(b\right)} $

Answer 1

i got the sam as you $$\frac{-9}{\log_a b}=-9\log_b a$$ which is not listed as Lord Shark the Unknown stated above