Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$

Question

Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$ We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$

Also the quadratic $6x^2-5x+1$ factors as $(2x-1)(3x-1)$. The equation then becomes $$\log_{1-2x}(2x-1)(3x-1)-\log_{1-3x}(2x-1)^2=2\\\log_{1-2x}(2x-1)(3x-1)-2\log_{1-3x}(1-2x)=2,$$ as $\log_{1-3x}(2x-1)^2=2\log_{1-3x}|2x-1|,$ but we know from $D_x$ that $2x-1<0$, $$\log_{1-2x}(2x-1)+\log_{1-2x}(3x-1)-\dfrac{2}{\log_{1-2x}(1-3x)}=2$$ I don't know what to do next.

Answer 1

You're almost there, but you may use a simplified method such as this:

$$\log_{1-2x}(2x-1)(3x-1) - 2\log_{1-3x}(1-2x)=2$$

$$\log_{1-2x}(1-2x)(1-3x) - 2\log_{1-3x}(1-2x)=2$$

$$\log_{1-2x}(1-2x)+\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=2$$ $$\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=1$$

$$\frac{1}{\log_{1-3x}(1-2x)} - 2\log_{1-3x}(1-2x)=1$$

Now you can set $\log_{1-3x}(1-2x)=t$ and get:

$$\frac{1}{t}-2t=1$$

This will give you a quadratic in terms of $t$ which you can solve for and substitute back into the substitution and solve for $x$ that way.

Answer 2

We have that for $1-2x>0$, $1-3x>0$, $1-2x\neq 1$, $1-3x\neq 1$

$$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2 $$

$$\iff \frac{\log((1-2x)(1-3x))}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=2$$

$$\iff \frac{\log (1-3x)}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=1$$

then by $u= \frac{\log (1-3x)}{\log (1-2x)}$ we obtain

$$u-\frac 2 u =1 \implies u = \frac{\log (1-3x)}{\log (1-2x)}=2 \implies x=\frac14$$

Answer 3

The key here is to use the change of base formula for logs: $\log_c(x)=\frac{\log_b(x)}{\log_b(c)}$. Using this allows us to express everything in terms of a common base, which then lets us use other log rules.

$$\frac{\log (1-2x)(1-3x)}{\log (1-2x)}-\frac{\log[(1-2x)^2]}{\log(1-3x)}=2.$$

Using log properties, this simplifies to

$$1+\frac{\log (1-3x)}{\log (1-2x)}-2\frac{\log[(1-2x)]}{\log(1-3x)}=2.$$

For ease of notation, let us write $u=\log(1-2x), v=\log(1-3x)$. Then $\frac{v}{u}-2\frac{u}{v}=1$. The easiest approach is to notice that we reciprocals, so that if $t=u/v$, then $1/t - 2t=1$. Multiplying through by $t$ yields a quadratic to solve for $t$, and then for $x$. However, if we do not make that realization, we may proceed as follows.

Multiplying by $uv$ to clear denominators, and moving everything to one side, we get

$$v^2-uv-2u^2=0$$

This factors as $$(v+u)(v-2u)=0$$ and so either $v+u=0$, or $v-2u=0$, i.e., either $\log(1-2x)(1-3x)=0$ so $(1-2x)(1-3x)=1$, or $\log\left(\frac{1-3x}{(1-2x)^2}\right)=0$, so $\frac{1-3x}{(1-2x)^2}=1$. This gives a linear and a quadratic equation to try to solve for $x$.