Question: Solve the equation
$$\log_3 \left(1 - 3x\right) = \log_9 \left(6x^{2} - 19x + 2 \right)$$
There's quite a bit going on, I'm trying to think about the best point to start in order to solve it.
The LHS can be expanded, the RHS has a quadratic in it, but the quadratic has a pretty nasty root that I don't think (?) will be of any use [1].
So as I'm solving for $x$ I need to try and isolate the $x$ terms... I don't really have a plan for this (I feel as though I should) so I'll just start and see.
I can expand the LHS to ;
$$\log_3 \left(1 - 3x\right)$$
$$ = \log_3 1 - \log_3 3x$$
Not too sure what step to take here, I know that $\log_3 3 = 1$ and that $\log_3 3x = \log_3 3 + \log_3 x$, I'm not sure how these help me with the solution at the moment though.
Seeing as the only thing I can think of is to get all terms isolated and move the $x$ to one side I'm just going to try and expand everything.
So the LHS first :
$$\log_3 \left(1 - 3x\right)$$
$$ = \log_3 1 - \log_3 3x$$
$$ = \log_3 1 - (\log_3 3 + \log_3 x)$$
$$ = \log_3 1 - \log_3 3 - \log_3 x$$
$$ = \log_3 1 - \log_3 3 - \log_3 x$$
$$ = 0 - 1 - \log_3 x$$
$$ = - 1 - \log_3 x$$
So that's the LHS expanded, now I'll do the same to the RHS.
As said previously, I don't think solving the quadratic in RHS helps as the numbers messy, so I'll just expand / simplify in a similar way to the LHS
$$\log_9 \left(6x^{2} - 19x + 2 \right)$$
$$\log_9 6x^2 - \log_9 19x + \log_9 2$$
This can be expanded further ;
$$2 \log_9 6x - (\log_9 19 + \log_9 x) + \log_9 2$$
$$ = 2 \log_9 6 + \log_9 x - \log_9 19 - \log_9 x + \log_9 2$$
$$ = 2 \log_9 6 - \log_9 19 + \log_9 2$$
Now all the terms are isolated I should be able to gather like terms (they're pretty much there anyway).
$$ -1 - \log_3 x= 2 \log_9 6 - \log_9 19 + \log_9 2$$
$$ - \log_3 x= 1 + 2 \log_9 6 - \log_9 19 + \log_9 2$$
Now simplify the RHS.
$$ 1 + 2 \log_9 6 - \log_9 19 + \log_9 2$$
$$ = 1 +\log_9 36- \log_9 19 + \log_9 2$$
Then using the multiplication rule
$$\log_9 19 + \log_9 2 $$
$$ = \log_9 (19*2)$$
$$ = \log_9 (38)$$
Which leaves the RHS at
$$1 + \log_9 36 - \log_9 38 $$
Using the division rule
$$1 + \log_9 36 - \log_9 38 $$
$$ = 1 + \log_9 \left(\frac{36}{38}\right)$$
$$ = 1 + \log_9 \left(\frac{18}{19}\right)$$
I'm not sure that this is the best expression to leave the RHS with or not at the mo \ldots{}
As it stands
$$ - \log_3 x = 1 + \log_9 \left(\frac{18}{19}\right)$$
So I can shift the LHS about a bit to try and get the $x$ term a little bit more isolated (pull the 3 out)
$$- \log_3 x $$
$$= - \left(\frac{\log x}{\log 3} \right)$$
Multiply both sides by $\log 3$
$$ - \log x= \log 3\left(1 + \log_9 \left(\frac{18}{19}\right)\right)$$
Pull the $1$ out
$$ - \log x= 2 \log 3\left( \log_9 \left(\frac{18}{19}\right)\right)$$
I'm going to expand out the fraction in the RHS so that I get all the terms in a common $\log$
$$\log_9 \left(\frac{18}{19}\right)$$
$$ = \log_9 18 - \log_9 19$$
$$ = \frac{\log 18}{\log 9} - \frac{\log 19}{\log 9}$$
$$ = \frac{\log 18 - \log 19}{\log 9}$$
Leaving the equation at
$$-\log x = 2 \log 3\left( \frac{\log 18 - \log 19}{\log 9}\right)$$
$$-\log x = \log 9\left( \frac{\log 18 - \log 19}{\log 9}\right)$$
Multiply the $\log 9$ through
$$-\log x = \log 18 - \log 19$$
Change the signs around
$$\log x = \log 19 - \log 18 $$
Therefore
$$\log x = \log 19 - \log 18 $$
$$ 10^{\left(\log 19 - \log 18\right)} = x$$
$$x = \frac{19}{18}$$
Which is way off the mark !
$$:( $$
The answer in the book is
$$-\frac{1}{3}, -2$$
This answer feels very convoluted, I'm not sure where I've gone wrong though. I'm inclined to think that it's a thinking error rather than dropping a digit somewhere, not too sure though. I'm self learning, so it's really helpful to get any input.
Thanks!
[1] Root of the quadratic in the RHS =
$$\frac{19 + \sqrt{313}}{12}$$
And
$$\frac{19 - \sqrt{313}}{12}$$
