How to solve the following inequality

${\mathrm{log}}_{{x}^{2}-3}({x}^{2}+6x)<{\mathrm{log}}_{x}(x+2)\text{}?$?

kadejoset
2022-07-17
Answered

How to solve ${\mathrm{log}}_{{x}^{2}-3}({x}^{2}+6x)<{\mathrm{log}}_{x}(x+2)$?

How to solve the following inequality

${\mathrm{log}}_{{x}^{2}-3}({x}^{2}+6x)<{\mathrm{log}}_{x}(x+2)\text{}?$?

How to solve the following inequality

${\mathrm{log}}_{{x}^{2}-3}({x}^{2}+6x)<{\mathrm{log}}_{x}(x+2)\text{}?$?

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Logarithmic inconsistency when integrating

Consider following integral:

$13\int \frac{1}{8x-4}dx$ (1)

By factorizing the denominator and then taking the factor outside the integral sign, it can be rewritten as

$\frac{13}{4}\int \frac{1}{2x-1}dx$ (2)

Now (1) and (2) should be equivalent, yet they evaluate into different integrals namely

$13\int \frac{1}{8x-4}dx=\frac{13}{8}\mathrm{ln}|8x-4|+C$ (1a)

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Since $\left(1\right)\equiv \left(2\right)$ , then (1a) and (2a) should be equivalent as well, which reduces to

$\mathrm{ln}|8x-4|=\mathrm{ln}|2x-1|$

which clearly isn't true. What am I missing here?

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