Minilog

Let $S$ be the expression. Then, we have:

$\begin{aligned} S & = 17\log_{30} x - 3 \log_x 5 + 20 \log_x 15 - 3 \log_x 6 + 20 \log_x 2 \\ & = \frac{17}{\log_x 30} + \log_x \left( \frac {15^{20} \dot{} 2^{20}} {5^3 \dot{} 6^3} \right) \\ & = \frac{17}{\log_x 30} + \log_x \left( 2^{17} \dot{} 3^{17} \dot{} 5^{17} \right) \\ & = \frac{17}{\log_x 30} + 17 \log_x 30 \quad \quad \color{#3D99F6} {[\text{by AM-GM inequality}]} \\ & \ge 2 \sqrt{\frac{17^2\log_x 30}{\log_x 30}} = \boxed{34} \end{aligned}$

Let $T(x)$ be the answer that log equation. so from log formula, we have

$T(x)=17 \log_{30}{x}-3 \log_{x}{(5)(6)}+20 \log_{x}{(15)(2)}\\ =17 \log_{30}{x}+17 \log_{x}{30} \\ = 17( \log_{30}{x}+\log_{x}{30}) \\ = 17( \log_{30}{x}+ \frac{1}{ \log_{30}{x}})$

Since $x>0$ then $\log_{30}{x}>0, \frac{1}{ \log_{30}{x}}>0.$

Use AM-GM : $T(x) \geq17({2 \sqrt{(log_{30}{x})(\frac{1}{ \log_{30}{x}})}}) \\ T(x) \geq{34}$ The minimum value $\boxed{34}$

The last part of the problem can be done as, Let $t=\log _{ x }{ 30 }$

Therefore, $S=17/t + 17t$ $S=17*{(1/t + t)}^{2} -34 >0$ So, we can say that, $S$ has minimum value of $34$