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प्रश्न
Solve for x.
log2 x + log4 x + log16 x = `21/4`
उत्तर
log2 x + log4 x + log16 x = `21/4`
∴ `logx/log2 + logx/log4 + logx/log16 = 21/4`
∴ `(logx)[1/log2 + 1/log4 + 1/log16] = 21/4`
∴ `(logx)[1/log2 + 1/log2^2 + 1/log2^4] = 21/4`
∴ `(logx) [1/log2 + 1/(2log2) + 1/(4log2)] = 21/4`
∴ `(logx/(log2)) [ 1 + 1/2 + 1/4] = 21/4`
∴ `(logx/(log2)) [(4 + 2 + 1)/4] = 21/4`
∴ `7/4.(logx/log2) = 21/4`
∴ `logx/log2` = 3
∴ log x = 3 log 2 = log23
∴ log x = log 8
∴ x = 8
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