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Let {a}>{1} be a real number. Then the number of roots equatio

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Let $a > 1$ be a real number. Then the number of roots equation $a^{2 (l o g)_{2} x} = 15 + 4 x^{(l o g)_{2} a}$ is 2 (b) infinite (c) 0 (d) 1

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Let

a > 1

be a real number. Then the number of roots equation

a^{2 (l o g)_{2} x} = 15 + 4 x^{(l o g)_{2} a}

is 2 (b) infinite (c) 0 (d) 1

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Text solutionVerified

Solution:

a^{2 l o g_{2}^{x}} = 15 + 4 x^{l o g_{2}^{a}}

a > 1

(a^{l o g_{2}^{x}})^{2} = 15 + 4 x^{l o g_{2}^{a}}

(x^{l o g_{2}^{a}})^{2} = 15 + 4 x^{l o g_{2}^{a}}

x > 0

Let

x^{l o g_{2}^{a}} = t

lo g_{2}^{a}

is positive

t^{2} = 15 + 4 t

t^{2} - 4 t - 15 = 0

t = \frac{4 \pm 16 + 60}{2} = \frac{4 \pm 76}{2}

t = 2 + 19

t = 2 - 19

this is not possible
Number of solution=1.

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Question Text	Let $a > 1$ be a real number. Then the number of roots equation $a^{2 (l o g)_{2} x} = 15 + 4 x^{(l o g)_{2} a}$ is 2 (b) infinite (c) 0 (d) 1
Answer Type	Text solution:1
Upvotes	150