Solving a nested logarithmic equation

[brotherbobby]

Homework Statement

Solve for $x$ : $\boldsymbol{\log_{x^2+x+1}\{\log_{2x^2+3x+5}(x^2+3)\}=0}$

Relevant Equations

1. Given $\log_b a= x$, we have the requirements that (1) $a>0$, (2) $b>0\; \text{and}\; b\ne 1$ and (3) that all $a,b,x \in \mathbb{R}$
2. If $\log_{f(x)} g(x) = 0\Rightarrow g(x) = 1\;\forall x$
3. If $\log_{f(x)} g(x) = 1\Rightarrow g(x) = f(x)\;\forall x$

Problem statement : Let me copy and paste the problem on the right as it appears in the text.

Solution : Using the Relevant Equations (2) and (3) above, we can claim that
\begin{align*}
&\log_{2x^2+3x+5}(x^2+3)=1\\
&\Rightarrow x^2+3 = 2x^2+3x+5\\
&\Rightarrow x^2+3x+2=0\\
&\Rightarrow (x+1)(x+2)=0\\
&\Rightarrow \underline{x = -1}\quad\text{OR}\quad \underline{x=-2}
&\end{align*}

We have three functions to consider for conditions satisfying positivity, as outlined in Relevant Equations (1) above, for the two (underlined) solutions just obtained.

1. $x^2+3$ : Both solutions satisfy this function being greater than zero.
2. $2x^2+3x+5$ : Likewise, both solutions satisfy the requirement of this "base" function for being greater than zero and not equal to one.
3. $x^2+x+1$ : For $x=-1$, this function is one, which is invalid. Hence this solution has to be discarded. However, for $x=-2$ this function is greater than zero and not one; so this solution holds good.

Answer : $\Large{\boxed{x = -2}}.$

Issue : The author says no answer for $x$ satisfies the logarithmic equation, giving $x=\varnothing$. I copy and paste his solution below :

Doubt : Is the author mistaken? Am I? A hint would be most welcome.

 

[Mark44]

Your answer looks fine to me. If x = -1, the outer log is base-1, which we can't have.
If x = -2, we have this:
$\log_{x^2 + x + 1}[\log_{2x^2 + 3x + 5}(x^2 + 3)] = \log_3[\log_7(7)] = \log_3(1) = 0$

I don't following the author's work at all, particularly where he writes $x \in R$ a couple of times, and then concludes that the solution set is empty.

 

[brotherbobby]

Your answer looks fine to me. If x = -1, the outer log is base-1, which we can't have.
If x = -2, we have this:
$\log_{x^2 + x + 1}[\log_{2x^2 + 3x + 5}(x^2 + 3)] = \log_3[\log_7(7)] = \log_3(1) = 0$

I don't following the author's work at all, particularly where he writes $x \in R$ a couple of times, and then concludes that the solution set is empty.

Thank you. Yes I checked my answer too. Matches the conditions. Sorry about the text.