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प्रश्न
Solve the following:
`log_2x + log_4x + log_16x = (21)/(4)`
उत्तर
`log_2x + log_4x + log_16x = (21)/(4)`
∴ `(1)/("log"_x2) + (1)/("log"_x2^2) + (1)/("log"_x2^4) = (21)/(4)`
∴ `(1)/("log"_x2) + (1)/("2log"_x2) + (1)/("4log"_x2) = (21)/(4)`
∴ `(1)/("log"_x2) (1 + 1/2 + 1/4) = (21)/(4)`
∴ `(1)/("log"_x2)(7/4) = (21)/(4)`
∴ `"log"_x2 = (7)/(4) . (4)/(21)`
∴ `"log"_x2 = (1)/(3)`
∴ `x^(1/3)` = 2
∴ x = 23
= 8.
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