Logarithm Basics Part I

$log_{ A }(x^{ 2 }+15)>log_{ A }(8x)\\ \Rightarrow \frac { log(x^{ 2 }+15) }{ logA } >\frac { log(8x) }{ logA }$

$\Rightarrow log(x^{ 2 }+15)<log(8x)$ Inequality reversed... think why ?

$\Rightarrow { x }^{ 2 }+15<8x\\ \Rightarrow { x }^{ 2 }-8x+15<0\\ \Rightarrow ({ x }-3)(x-5)<0$

Now, any $x$ less than $3$ would make both $(x-3)$ and $(x-5)$ negative. Thus their product would be positive and hence not less than $0$ . Also, any $x$ greater than $5$ would make their product to be positive and hence not less than $0$ . we also can't have $x$ as $3$ or $5$ since that would make the product $0$ where as the conditions says strictly less than.

Only an $x$ between $3$ and $5$ would make their product negative.

Hence, $x$ belongs to $(3,5)$

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Allow me to expand upon Vaibhav's solution. If we take into account the triangle assumption provided to us, then for any $z=a+bi (a,b \in \mathbb{R})$ :

$z^3 + iz^2 + 2i = 0 \Rightarrow (a+bi)^3 + i(a+bi)^2 + 2i = 0$ ;

or $(a^3 +3a^2bi - 3ab^2 -b^3i) + (a^2-b^2+2abi)i + 2i = 0;$

or $a^3 - 3ab^2 - 2ab = 0$ AND $3a^2b - b^3 + a^2 - b^2 + 2 = 0$ ;

or $(a,b) = (\pm 1, -1); (0,1)$ .

Thus, $A = \frac{1}{3} \cdot |\frac{1}{2} \cdot \begin{vmatrix} 1 & 0 & 1 \\1 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix}| = \frac{2}{3}$ , which gives us the inequality $\log_{2/3}(x^2+15) > \log_{2/3}(8x) \Rightarrow \frac{\log(x^2+15)}{\log(2/3)} > \frac{\log(8x)}{\log(2/3)}$ . Since $\log(2/3) < 0$ , we require a change of direction in the inequality, or $\log(x^2+15) < \log(8x)$ . This now gives us:

$x^2 + 15 < 8x \Rightarrow x^2 - 8x + 15 <0 \Rightarrow (x-3)(x-5) < 0 \Rightarrow \boxed{x \in (3,5)}$ .