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We have a^{x} = N, then _{a}N = x and _{2}ab = _{2}a+ _{2}b. T

Question

We have $a^{x} = N$ , then $lo g_{a} N = x$ and $lo g_{2} ab = lo g_{2} a + lo g_{2} b$ . Then,
The number of solutions of the two given equations $lo g_{2} (x y) = 5$ and $lo g_{0.5} (\frac{x}{y}) = 1$ is

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Amelia Discussed

We have

a^{x} = N

, then

lo g_{a} N = x

and

lo g_{2} ab = lo g_{2} a + lo g_{2} b

. Then,

The number of solutions of the two given equations

lo g_{2} (x y) = 5

and

lo g_{0.5} (\frac{x}{y}) = 1

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Text solutionVerified

lo g_{2} (x y) = 5

and

lo g_{0.5} (\frac{x}{y}) = 1

∴ x y = 32

and

\frac{x}{y} = \frac{1}{2}

\Rightarrow x y \times \frac{x}{y} = 16 \Rightarrow x^{2} = 16 \Rightarrow x = \pm 4

Substituting the value of x in the equation

y = 2 x

y = \pm 8

Thus (4, 8) and (-4, -8) are the possible solutions.

Hence the number of solutions

= 2

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Question Text	We have $a^{x} = N$ , then $lo g_{a} N = x$ and $lo g_{2} ab = lo g_{2} a + lo g_{2} b$ . Then, The number of solutions of the two given equations $lo g_{2} (x y) = 5$ and $lo g_{0.5} (\frac{x}{y}) = 1$ is
Answer Type	Text solution:1
Upvotes	150