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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 ?

SOORAJ SOMAN
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3 Answers3

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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.

Allawonder
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  • 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 – SOORAJ SOMAN Apr 27 '19 at 08:40
  • @ss1729 Turns out I'd even made a mistake. Contrary to what I'd claimed, $-1$ is not a root of the equation. I missed a minus somewhere. In any case I have put up a different answer. Have a look and see if it sates your curiosity. – Allawonder Apr 27 '19 at 09:29
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    The fourth degree polynomial should be $(2x+3)^3(5x+4)-1$ – Lozenges Apr 27 '19 at 09:45
  • @Lozenges Yes, that's what it is , if you expand it out. – Allawonder Apr 27 '19 at 12:05
  • could u please comment on my edit in OP ? – SOORAJ SOMAN Apr 27 '19 at 16:38
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    I think you have expanded $(2x+3)(5x+4)^3-1$ – Lozenges Apr 27 '19 at 18:12
  • ok. I think ur post states there must be a minimum value between $0$ and $-1$. But wht abt the solution? Could u please make it a little more rigorous ? – SOORAJ SOMAN Apr 27 '19 at 19:30
  • @Lozenges Wow, you were right! I just realised the slip now. I don't understand why my brain does these little permutations without my permission. Thank you. PS. I have now rewritten my answer. – Allawonder Apr 28 '19 at 03:52
  • @ss1729 I had made another slip -- I had been dealing with another polynomial. Now I have rewritten the answer. And it's no longer a sketch, but a complete proof that there is no root of that equation satisfying the given conditions. – Allawonder Apr 28 '19 at 03:53
  • So u suggesting there is no root for the second case ?. Please check @egreg post seems to suggest there is one in the second case. – SOORAJ SOMAN Apr 28 '19 at 06:51
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    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. – SOORAJ SOMAN Apr 28 '19 at 07:17
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    @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. – Allawonder Apr 28 '19 at 17:35
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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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egreg
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  • I have edited OP please do check. – SOORAJ SOMAN Apr 27 '19 at 11:34
  • 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 ! – SOORAJ SOMAN Apr 27 '19 at 16:58
  • @ss1729 No, just one, because the other one doesn't respect the restrictions on $x$. – egreg Apr 27 '19 at 19:24
  • 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 . – SOORAJ SOMAN Apr 27 '19 at 19:31
  • @ss1729 There is a formula for degree 4 equations. If you try it at WolframAlpha, you’ll get the explicit form. – egreg Apr 27 '19 at 20:37
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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

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

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

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

giuseppe mancò
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