# Basic Logarithm equation, and how best to approach this question logically

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

$$\frac{19 - \sqrt{313}}{12}$$

• Here is one mistake $\log_{3}(1 - 3x) \neq \log_{3}(1) - \log_{3}(3x)$. $\log(a - b) = \log(a) - \log(b)$ is not a rule. – user222031 May 15 '15 at 0:27
• As pointed out by @user222031, your first line is wrong. Perhaps this would help. – David May 15 '15 at 0:29
• @user222031 ah damnit, nice one :) – baxx May 15 '15 at 0:31
• @David thanks for the link, – baxx May 15 '15 at 0:31

Make use of the fact that $$\log_a(y) = \log_{a^2}\left(y^2 \right)$$ We hence have that $$\log_3(1-3x) = \log_9\left((1-3x)^2\right)$$ We hence have $$(1-3x)^2 = 6x^2-19x+2 \implies 9x^2 - 6x+1 = 6x^2-19x+2 \implies 3x^2 + 13x - 1 =0$$ Hence, we have $$x= \dfrac{-13 \pm \sqrt{181}}6$$