adsense
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.
The answer:
Select REMOVE
Condition: \(x > 0\).
We have: \({\log _3}\left( {{x^2} + {y^2} + x} \right) + {\log _2}\left( {{x^2} + {y^2}} \right) \le {\log _3}x + {\log _2}\left( {{x^2} + {y^2} + 24x} \right)\)
\( \Leftrightarrow {\log _3}\left( {{x^2} + {y^2} + x} \right) – {\log _3}x \le {\log _2}\left( {{x) ^2} + {y^2} + 24x} \right) – {\log _2}\left( {{x^2} + {y^2}} \right)\)
\( \Leftrightarrow {\log _3}\left( {\frac{{{x^2} + {y^2} + x}}{x}} \right) \le {\log _2}\left( { { \frac{{{x^2} + {y^2} + 24x}}{{{x^2} + {y^2}}}} \right)\)\( \Leftrightarrow {\log _3}\ left( {1 + \frac{{{x^2} + {y^2}}}{x}} \right) \le {\log _2}\left( {1 + \frac{{24x}}{ {{x^2} + {y^2}}}} \right)\)
\( \Leftrightarrow {\log _3}\left( {\frac{{{x^2} + {y^2}}}{x} + 1} \right) – {\log _2}\left( {1 + \frac{{24x}}{{{x^2} + {y^2}}}} \right) \le 0. {\rm{ }}\)
adsense
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.
