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 ?

  • $\begingroup$ @DonAntonio how is that possible ? $\endgroup$ – ss1729 Apr 27 at 7:55
  • $\begingroup$ @ss It actually is not: I overlooked a factor there. $\endgroup$ – DonAntonio Apr 27 at 7:57
  • $\begingroup$ The original equation has a real root $\approx -0.7425$. Why is that not acceptable? $\endgroup$ – Lozenges Apr 27 at 9:11
  • $\begingroup$ Just noticed something: $\log _ba =\frac{\log a}{\log b}$. You have written the reciprocal. It does not matter in case 1 but in case 2 you get another solution ≈−0.7425 $\endgroup$ – Lozenges Apr 27 at 9:25
  • $\begingroup$ @Lozenges thnx for mentioning, it was by mistake. $\endgroup$ – ss1729 Apr 27 at 10:30

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.

  • $\begingroup$ I think u r suggesting the rational root theorem, to get clues abt the roots of a 4th order polynomial, but it is bit cumbersome, looking for a better way $\endgroup$ – ss1729 Apr 27 at 8:40
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    $\begingroup$ The fourth degree polynomial should be $(2x+3)^3(5x+4)-1$ $\endgroup$ – Lozenges Apr 27 at 9:45
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    $\begingroup$ I think you have expanded $(2x+3)(5x+4)^3-1$ $\endgroup$ – Lozenges Apr 27 at 18:12
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    $\begingroup$ I think $p(-0.8)=-1<0$ not greater than 0 as u are suggesting, if thats the case given $p(-0.6)>0$ there must be a root in the permitted interval. $\endgroup$ – ss1729 Apr 28 at 7:17
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    $\begingroup$ @ss1729 My slovenliness with manual calculation has caught me out again. You're right (had to check with an online calculator). Thus, it seems the conclusion of my proof should have been that indeed there is a root in the permitted interval, so that the claim in the text is incorrect. I shall now change this error in my answer. Thank you. $\endgroup$ – Allawonder Apr 28 at 17:35

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.

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  • $\begingroup$ I have edited OP please do check. $\endgroup$ – ss1729 Apr 27 at 11:34
  • $\begingroup$ So are u suggesting it has got two solutions in the second case ?. I am bit confused as my reference and @Allawonder suggesting there is no solution ! $\endgroup$ – ss1729 Apr 27 at 16:58
  • $\begingroup$ @ss1729 No, just one, because the other one doesn't respect the restrictions on $x$. $\endgroup$ – egreg Apr 27 at 19:24
  • $\begingroup$ Alright. But I was wondering is there an analytical way to find the exact root, as this was asked as a multiple choice question ?. and obviously my reference seems to be wrong about the real roots . $\endgroup$ – ss1729 Apr 27 at 19:31
  • $\begingroup$ @ss1729 There is a formula for degree 4 equations. If you try it at WolframAlpha, you’ll get the explicit form. $\endgroup$ – egreg Apr 27 at 20:37

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