Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$

Question

Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$

My Attempt $$ 2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5 $$ $$ 3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x+3)=1\\ 3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-\log_{(2x+3)}(2x+3)=1\\ 3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-1=1\\ 3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)=2\\ 3\log_{(5x+4)}(2x+3)-\frac{1}{\log_{(5x+4)}(2x+3)}=2 $$ Set $y=\log_{(5x+4)}(2x+3)$ $$ 3y^2-2y-1=0\implies y=\log_{(5x+4)}(2x+3)=1,\frac{-1}{3} $$ Case 1: $$ \log_{(5x+4)}(2x+3)=\frac{\log(5x+4)}{\log(2x+3)}=1\implies\log(5x+4)=\log(2x+3) \implies \boxed{x=\frac{-1}{3}} $$ Case 2: $$ \log_{(5x+4)}(2x+3)=\frac{\log(2x+3)}{\log(5x+4)}=\frac{-1}{3}\implies \color{red} ? $$ My reference says $x=\frac{-1}{3}$ is the only solution. How do I prove it ?

Possible Solution $$ y=\log_{(5x+4)}(2x+3)=\frac{-1}{3}<0\:,\quad x>-4/5=-0.8 $$ Case 1: $5x+4>1\implies x>-3/5=-0.6$ $$ 0<2x+3<1\implies -3/2<x<-1\implies-1.5<x<-1\\ \text{Not Possible !} $$ Case 2: $0<5x+4<1\implies -4/5<x<-3/5\implies-0.8<x<-0.6$ $$ 2x+3>1\implies x>-1\\ \implies \boxed{x\in(-0.8,-0.6)} $$ It seems like there could be a solution for "case 2" in my attempt ?

Answer 1

One can show why this case yields exactly one solution, as follows.

First, the expression on $\text{LHS}$ of the equation $$\log_{(5x+4)}{(2x+3)}=-\frac 13$$ uniquely defines a real number provided that $1\ne 5x+4>0$ and $2x+3>0.$ Thus, any solution of the equation must also satisfy the conditions $$x>-0.8, x\ne -0.6.$$

Having done that, I now show that this equation has exactly one root satisfying these conditions, thereby confirming the claim.

This equation is, by definition, equivalent to the polynomial equation $$40x^4+212x^3+414x^2+351x+107=p(x)=0.$$ First, note that this has no real roots for $x\ge 0$ since then $p(x)>0.$ Now since $p(-1)<0,$ it follows that there is at least one root in $(-1,0).$ We show that there is in fact a unique root in this interval. To see this, take the first derivative of $p(x),$ which gives $$p'(x)=160x^3+636x^2+828x+351.$$ The second derivative is then given by $$p''(x)=40x^2+106x+69,$$ whose discriminant is $14,$ whence its roots are $-1.5,-1.15,$ both outside of the interval $(-1,0).$ It follows that $p''>0$ in this interval. Therefore, the first derivative $p'(x)$ increases for all $x\in(-1,0).$ This means $p$ is convex in that interval; together with the negativity of $p$ at $-1,$ it implies that $p$ has at most one root in this interval. Finally, I show that this root must lie in $(-0.8,-0.6),$ finishing off the problem.

Now, we have that $p(-0.6)>0.$ Furthermore, we have that $p(-0.8)<0.$ Thus, we have confirmed the claim that the unique root in $(-1,0)$ falls within the subinterval $(-0.8,-0.6),$ which lies within the permitted range $-0.6\ne x>-0.8.$ Since there are no roots for $x\ge -0.6,$ it follows that no other root of $p(x)=0$ falls in the permitted range. This completes the proof.

Answer 2

Since $10x^2+23x+12=(2x+3)(5x+4)$, my suggestion is to set $t=\log_{(2x+3)}(5x+4)$, so that $$ \log_{(5x+4)}(2x+3)=\frac{1}{t} $$ and the equation becomes $$ \frac{3}{t}-1-t=1 $$ so $t^2+2t-3=0$ and $t=1$ or $t=-3$.

The first case yields $5x+4=2x+3$, hence $x=-1/3$.

The second case yields $5x+4=(2x+3)^{-3}$. If we consider the two functions $f(x)=5x+4$ and $g(x)=(2x+3)^{-3}$, we can see that $$ f(-2)=-6,\quad g(-2)=-1,\qquad f(-1)=-1,\quad g(-1)=1,\qquad f(0)=4,\quad g(0)=1/27 $$ Hence the equation $f(x)=g(x)$ has two roots in the interval $(-2,0)$. For the one in the interval $(-2,-1)$, we have $f(x)=g(x)<0$, so this is not a solution.

For the one in the interval $(-1,0)$, we have $f(x)=g(x)>0$ and not $1$. So the solution is good for the problem.

Answer 3

The only permissible root, coming from the polynomial:

$40x^{4}+212x^{3}+414x^{2}+351x+107=0$,

along with $x=\frac{1}{3}$, is as follows:

$x=\sqrt{-S+\sqrt{R}}+\sqrt{T}-i$,

where $R$, $S$ and $T$ are given by:

$R=-A^{1/3}+B^{1/3}+a^{1/3}+b^{1/3}+f$,

$S=C^{1/3}-c^{1/3}-g$,

$T=D^{1/3}-d^{1/3}+h$;

and again

$A=\frac{ 117649\sqrt{386481}}{ 294912000000000}-\frac {5764801}{ 32768000000000}$, $a=\frac{ 117649\sqrt{386481}}{ 294912000000000}+\frac {5764801}{ 32768000000000}$; $B=\frac{96827}{55296000000}-\frac{49\sqrt{386481}}{18432000000}$, $b=\frac{96827}{55296000000}+\frac{49\sqrt{386481}}{18432000000}$;

$C=\frac{\sqrt{386481}}{4608000}-\frac{49}{512000}$, $c=\frac{\sqrt{386481}}{4608000}+\frac{49}{512000}$;

$D=C$, $d=c$; $f=\frac{23203}{1920000}$, $g=\frac{49}{800}$, $h=\frac{49}{1600}$, $i=\frac{53}{40}$.