JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 19)
If $${\log _3}2,{\log _3}({2^x} - 5),{\log _3}\left( {{2^x} - {7 \over 2}} \right)$$ are in an arithmetic progression, then the value of x is equal to _____________.
Answer
3
Explanation
$$2{\log _3}({2^x} - 5) = {\log _2} + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$$
Let $${2^x} = t$$
$${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$$
$${(t - 5)^2} = 2t - 7$$
$${t^2} - 12t + 32 = 0$$
$$(t - 4)(t - 8) = 0$$
$$\Rightarrow$$ 2x = 4 or 2x = 8
x = 2 (Rejected)
Or x = 3
Let $${2^x} = t$$
$${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$$
$${(t - 5)^2} = 2t - 7$$
$${t^2} - 12t + 32 = 0$$
$$(t - 4)(t - 8) = 0$$
$$\Rightarrow$$ 2x = 4 or 2x = 8
x = 2 (Rejected)
Or x = 3
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