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I tried so hard to solve this, but I've always failed, can someone solve it for me and clarify the steps please?

$\log_x(x+2) < \log_{x+5}(x+2) + \log_5(x+3)$.

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  • $\begingroup$ Thank you, I'll check that link. But I guess that my question was clear enough, right? $\endgroup$ – Mrkinix Mar 28 at 13:19
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    $\begingroup$ Quite clear, yes. But please check that link first and then tell me if you think that your question really follows the suggestions provided there. $\endgroup$ – José Carlos Santos Mar 28 at 13:25
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If you use the change of base property, letting $ln$ be the natural log, you have:

$$\frac{ln(x + 2)}{ln(x + 5)} + \frac{ln(x + 3)}{ln(5)} > \frac{ln(x + 2)}{ln(x)}$$

which clearly holds for $0$ $<$ $x$ $<$ $1$, since only r.h.s. being negative in this interval.

However, further working out the expression, it can be obtained that:

$$\frac{-ln(x + 2)ln(x + 5)ln(5) + ln(x + 2)ln(x)ln(5) + ln(x + 3)ln(x + 5)ln(x)}{ln(x)ln(x + 5)ln(5)} > 0$$

which holds not only for $0$ $<$ $x$ $<$ $1$ (both nominator and denominator negative), but also if $x$ $>$ $2.27745$, which instead makes both nominator and denominator positive.

You can try to derive a closed form expression for the latter case, which I instead found using Matlab. It is also possible that I am missing some more elegant way to solve it using a different change of base.

Hope it helps!

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