Question
If log₂(x) + log₄(x) + log₈(x) = 11, with x > 0,
then the value of x is:Solution
ATQ, We convert all logarithms to the same base (base 2). Note that: 4 = 2², so log₄(x) = log₂(x) / log₂(4) = log₂(x) / 2 8 = 2³, so log₈(x) = log₂(x) / log₂(8) = log₂(x) / 3 Let y = log₂(x). Then the equation becomes: y + (y/2) + (y/3) = 11 Find a common denominator (6): (6y/6) + (3y/6) + (2y/6) = 11 (11y/6) = 11 So: 11y/6 = 11 y = 11 * (6/11) y = 6 Recall y = log₂(x), So: log₂(x) = 6 x = 2⁶ = 64 Therefore, x = 64
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