JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 2)
A possible value of 'x', for which the ninth term in the expansion of $${\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}} \right\}^{10}}$$ in the increasing powers of $${3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}$$ is equal to 180, is :
0
$$-$$1
2
1
Explanation
$${}^{10}{C_8}({25^{(x - 1)}} + 7) \times {({5^{(x - 1)}} + 1)^{ - 1}} = 180$$
$$ \Rightarrow {{{{25}^{x - 1}} + 7} \over {{5^{(x - 1)}} + 1}} = 4$$
$$ \Rightarrow {{{t^2} + 7} \over {t + 1}} = 4$$;
$$\Rightarrow$$ t = 1, 3 = 5x $$-$$ 1
$$\Rightarrow$$ x $$-$$ 1 = 0 (one of the possible value).
$$\Rightarrow$$ x = 1
$$ \Rightarrow {{{{25}^{x - 1}} + 7} \over {{5^{(x - 1)}} + 1}} = 4$$
$$ \Rightarrow {{{t^2} + 7} \over {t + 1}} = 4$$;
$$\Rightarrow$$ t = 1, 3 = 5x $$-$$ 1
$$\Rightarrow$$ x $$-$$ 1 = 0 (one of the possible value).
$$\Rightarrow$$ x = 1
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