Discussion

Question:

$\log_{x^{2}} (y)$ + $\log_{y^{2}} (x)$ = 1 and y = x² - 30, then the value of x² + y² is:

Explanation:

Given, $\log_{x^{2}} (y)$ + $\log_{y^{2}} (x)$ = 1

⇒ $\frac{\log_{x} (y)}{2}$ + $\frac{\log_{y} (x)}{2}$ = 1

⇒ log_xy + log_yx = 2

Now, log_xy and log_yx are reciprocal of each other and their sum is 2. Sum of a number and its reciprocal is always greater than or equal to 2. Equality is true when the number and reciprocal are both equal to 1.

∴ log_xy = 1
⇒ x = y

Now, y = x² - 30 [y = x]
⇒ x² - x - 30 = 0
⇒ (x - 6)(x + 5) = 0
⇒ x = 6 [x = -5 is rejected as log is not defined for negative numbers.]
∴ y = 6

∴ x² + y² = 6² + 6² = 72

Hence, 72.

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Doubts

RIDDHI GUPTA said (2024-04-23 06:03:22)

WHY IS THE SUM OF NO AND ITS RECIPROCAL IS ALWAYS EQUAL OR GREATER THAN 2

Reply from Admin:

Let the number be x and its reciprocal be 1/x.

We know Arithmetic Mean ≥ Geometric Mean. Applying this for x and 1/x

⇒ AM ≥ GM
⇒ (x + 1/x)/2 ≥ √(x × 1/x)
⇒ (x + 1/x)/2 ≥ √1
⇒ (x + 1/x)/2 ≥ 1
⇒ x + 1/x ≥ 2

∴ Sum of a positive number and its reciproal is always greater than or equal to 0.

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