Consider the function $y = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } }$, where $x \geqq 0$. (1) We are to find the $x$ at which $y$ is minimized. When we differentiate $y$, we have $$\frac { d y } { d x } = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } \left( 2 x \log _ { e } \mathbf { A } - \mathbf { B } \log _ { e } \mathbf { C } \right) .$$ Hence, when we express the value of $x$ at which $y$ is minimized using the common logarithm, we have $$x = \frac { \mathbf { D } } { \mathbf { F } \left( 1 - \log _ { 10 } \mathbf { E } \right) } .$$ (2) We are to find the smallest positive integer $x$ satisfying $\frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } > 1000$. From the inequality $y > 1000$, we obtain $$x ^ { \mathbf { H } } \log _ { 10 } \mathbf { I } - \mathbf { J } x \log _ { 10 } \mathbf { L } - \mathbf { K } > 0 .$$ When we solve the inequality using $0.3$ as an approximate value for $\log _ { 10 } 2 = 0.301 \cdots$, the smallest positive integer $x$ satisfying $y > 1000$ is $\mathbf{Q}$.
Consider the function $y = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } }$, where $x \geqq 0$.
(1) We are to find the $x$ at which $y$ is minimized.
When we differentiate $y$, we have
$$\frac { d y } { d x } = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } \left( 2 x \log _ { e } \mathbf { A } - \mathbf { B } \log _ { e } \mathbf { C } \right) .$$
Hence, when we express the value of $x$ at which $y$ is minimized using the common logarithm, we have
$$x = \frac { \mathbf { D } } { \mathbf { F } \left( 1 - \log _ { 10 } \mathbf { E } \right) } .$$
(2) We are to find the smallest positive integer $x$ satisfying $\frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } > 1000$.
From the inequality $y > 1000$, we obtain
$$x ^ { \mathbf { H } } \log _ { 10 } \mathbf { I } - \mathbf { J } x \log _ { 10 } \mathbf { L } - \mathbf { K } > 0 .$$
When we solve the inequality using $0.3$ as an approximate value for $\log _ { 10 } 2 = 0.301 \cdots$, the smallest positive integer $x$ satisfying $y > 1000$ is $\mathbf{Q}$.