## How many pairs of integers (x;y) satisfy ({log _3}left( {{x^2} + {y^2} + x} right) + {log _2}left( {{x^2} ) + {y^2}} right) le {log _3}x + {log _2}left( {{x^2} + {y^2} + 24x} right)?) – Math Book

How many pairs of integers are there? $$(x;y)$$ satisfy

$${\log _3}\left( {{x^2} + {y^2} + x} \right) + {\log _2}\left( {{x^2} + {y^2}} \right) \le {\log _3}x + {\log _2}\left( {{x^2} + {y^2} + 24x} \right)?$$

A. 89.

B. 48.

C. 90.

D. 49.

Select REMOVE

Condition: $$x > 0$$.

We have: $${\log _3}\left( {{x^2} + {y^2} + x} \right) + {\log _2}\left( {{x^2} + {y^2}} \right) \le {\log _3}x + {\log _2}\left( {{x^2} + {y^2} + 24x} \right)$$

$$\Leftrightarrow {\log _3}\left( {{x^2} + {y^2} + x} \right) – {\log _3}x \le {\log _2}\left( {{x) ^2} + {y^2} + 24x} \right) – {\log _2}\left( {{x^2} + {y^2}} \right)$$

$$\Leftrightarrow {\log _3}\left( {\frac{{{x^2} + {y^2} + x}}{x}} \right) \le {\log _2}\left( { { \frac{{{x^2} + {y^2} + 24x}}{{{x^2} + {y^2}}}} \right)$$$$\Leftrightarrow {\log _3}\ left( {1 + \frac{{{x^2} + {y^2}}}{x}} \right) \le {\log _2}\left( {1 + \frac{{24x}}{ {{x^2} + {y^2}}}} \right)$$

$$\Leftrightarrow {\log _3}\left( {\frac{{{x^2} + {y^2}}}{x} + 1} \right) – {\log _2}\left( {1 + \frac{{24x}}{{{x^2} + {y^2}}}} \right) \le 0. {\rm{ }}$$

Put: $$t = \frac{{{x^2} + {y^2}}}{x}(t > 0)$$the inequality becomes: $${\log _3}(1 + t) – {\log _2}\left( {1 + \frac{{24}}{t}} \right) \le 0$$ (first).

Consider the function $$f Derive a covariate function on the interval \((0; + \infty )$$.

We have $$f(8) = {\log _3}(1 + 8) – {\log _2}\left( {1 + \frac{{24}}{8}} \right) = 0$$

Thence inferred: $$(1) \Leftrightarrow f Count pairs of integer values ​​of \((x;y)$$

We have: $${(x – 4)^2} \le 16 \Leftrightarrow 0 \le x \le 8$$but $$x > 0$$ should $$0 < x \le 8$$.

With $$x = 1,x = 7 \Rightarrow y = \{ \pm 2; \pm 1;0\}$$ there should be 10 pairs.

With $$x = 2,x = 6 \Rightarrow y = \{ \pm 3; \pm 2; \pm 1;0\}$$ there should be 14 pairs.

With $$x = 3,x = 5 \Rightarrow y = \{ \pm 3; \pm 2; \pm 1;0\}$$ there should be 14 pairs.

With $$x = 4 \Rightarrow y = \{ \pm 4; \pm 3; \pm 2; \pm 1;0\}$$ there should be 9 pairs.

With $$x = 8 \Rightarrow y = 0$$ have 1 pair.

So there are 48 pairs of integer values $$(x;y)$$ satisfy the topic.