Very strange logarithm simplification

Question

I have very strange logarithm simplification:

$$\begin{array}{l}\frac{{{{\log }_{{2^{{{(x + 1)}^2}}} - 1}}({{\log }_{2{x^2} + 2x + 3}}({x^2} - 2x))}}{{{{\log }_{{2^{{{(x + 1)}^2}}} - 1}}({x^2} + 6x + 10)}} \ge 0 \Leftrightarrow {\log _{{x^2} + 6x + 10}}{\log _{2{x^2} + 2x + 3}}({x^2} - 2x) \ge 0\\\\when\,x \ne - 3,x \ne - 2,x \ne - 1,x \ne 0\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaWaaWbaaWqabeaa % caGGOaGaamiEaiabgUcaRiaaigdacaGGPaWaaWbaaeqabaGaaGOmaa % aaaaWccqGHsislcaaIXaaabeaakiaacIcaciGGSbGaai4BaiaacEga % daWgaaWcbaGaaGOmaiaadIhadaahaaadbeqaaiaaikdaaaWccqGHRa % WkcaaIYaGaamiEaiabgUcaRiaaiodaaeqaaOGaaiikaiaadIhadaah % aaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamiEaiaacMcacaGGPa % aabaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdadaahaaadbeqa % aiaacIcacaWG4bGaey4kaSIaaGymaiaacMcadaahaaqabeaacaaIYa % aaaaaaliabgkHiTiaaigdaaeqaaOGaaiikaiaadIhadaahaaWcbeqa % aiaaikdaaaGccqGHRaWkcaaI2aGaaiiEaiabgUcaRiaaigdacaaIWa % GaaiykaaaacqGHLjYScaaIWaGaeyi1HSTaciiBaiaac+gacaGGNbWa % aSbaaSqaaiaadIhadaahaaadbeqaaiaaikdaaaWccqGHRaWkcaaI2a % GaaiiEaiabgUcaRiaaigdacaaIWaaabeaakiGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaGaamiEamaaCaaameqabaGaaGOmaaaaliabgU % caRiaaikdacaWG4bGaey4kaSIaaG4maaqabaGccaGGOaGaamiEamaa % CaaaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG4bGaaiykaiabgw % MiZkaaicdaaeaaaeaacaGG3bGaaiiAaiaacwgacaGGUbGaaGPaVlaa % cIhacqGHGjsUcqGHsislcaaIZaGaaiilaiaacIhacqGHGjsUcqGHsi % slcaaIYaGaaiilaiaacIhacqGHGjsUcqGHsislcaaIXaGaaiilaiaa % cIhacqGHGjsUcaaIWaaaaaa!9DAE! $$

Is it valid?

For me it's the same as

$$\frac{{{{\log }_a}{{\log }_b}c}}{{{{\log }_a}d}} \ge 0 \Leftrightarrow {\log _d}{\log _b}c \ge 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGSbGaai4BaiaacEgadaWgaaWcbaGaamyyaaqabaGcciGGSbGaai4B % aiaacEgadaWgaaWcbaGaamOyaaqabaGccaWGJbaabaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadggaaeqaaOGaamizaaaacqGHLjYScaaI % WaGaeyi1HSTaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadsgaaeqaaO % GaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadkgaaeqaaOGaam4yaiab % gwMiZkaaicdaaaa!53BC! $$

I can't understand which rules leads to this simplification.

Answer 1

You have $$ \frac{\log_a \log_b c}{\log_a d} = \log_d \log_b c $$ by well known rule $$ \log_d X = \frac{\log_a X}{\log_a d} $$ applied with $X=\log_b c$.

So the two expressions you're comparing to $0$ are equal, and so of course they are nonnegative at the same time.

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You need more conditions than $x\notin\{-3,-2,-1,0\}$, though. In order for the logarithm of $x^2-2x$ to exist and be positive, we must have $x^2-2x>1$, which excludes the closed interval $[1-\sqrt2,1+\sqrt2]$.