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 If log2X + log4X = log0.25√6 and x > 0, then x is
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 If log2X + log4X = log0.25√6 and x > 0, then x is Related: Logarithm...
Solution:

Given log2X log4X = log0.25√6 and x > 0

We know that,

loga (x) loga (y) = loga (x*y)

Therefore,

log2X log4X = log2X*4X = log8X^2

Now the given equation can be written as:

log8X^2 = log0.25√6

Converting RHS to base 8:

log8 (0.25√6) = log8 (0.25) + log8 (√6) = -1 + log8 (√6)

Therefore,

log8 X^2 = -1 + log8 (√6)

Taking antilog of both the sides:

X^2 = 8^(-1) * (√6)^8

X^2 = (√6)^(-1)

X = 1/√6

Hence, the value of x is 1/√6.
Community Answer
 If log2X + log4X = log0.25√6 and x > 0, then x is Related: Logarithm...
Log2x+log4x=log0.25√6
[log(a×b) =loga+logb]
log(2x×4x) =log0.25√6
from antilog-
2x×4x=0.25√6
8x=0.25√6
x=25√6÷800
x=√6÷32
x=0.0765
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 If log2X + log4X = log0.25√6 and x > 0, then x is Related: Logarithms - Important Formulas, Quantitative Aptitude?
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