Sainik School Entrance Exam 2020 Solved Paper

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Question : 26 of 150

Marks: +1, -0

Find the value of

x: 3^{2 x} ×3^{x+3} ×3^{4-x}=(√{3})^{10}

$-1$
0
1
2

Solution:

(a)

3^{2 x} ×3^{x+3} ×3^{4-x}=(√{3})^{10}

⇒3^{2 x+x+3+4-x}=[(3)^{1 / 2}]^{10}

⇒3^{2 x+7}=3^5

...........[∵ base is equal hence exponent is equal]

⇒2 x+7=5

⇒2 x=5-7=-2

⇒x=-{2}/{2}=-1

∴

x=-1

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