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प्रश्न
Solve log8x + log4x + log2x = 11
उत्तर
log8x = `1/log_x 8 = 1/log_x 2^3 = 1/(3log_x 2)`
log4x = `1/log_x 4 = 1/log_x 2^2 = 1/(2log_x 2)`
log2x = `1/log_x 2`
∴ log8x + log4x + log2x = 11
⇒ `1/(3log_x 2) + 1/(2log_x 2) + 1/log_x 2` = 11
⇒ `1/log_x 2 [1/3 + 1/2 + 1]` = 11
⇒ `1/log_x 2[(2 + 3 + 6)/6]` = 11
⇒ `1/log_x 2 = 11 xx 6/11` = 6
(i.e.,) = 6
x = 26
= 64
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