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- 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##

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

**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.

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