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2 May, 21:51

Solve for b if: 4log base2 2x + 2log base b x = 4

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  1. 3 May, 00:03
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    Log base b=logb

    4 log₂ 2x + 2logb x=4

    log₂ (2x) ⁴+logb x²=log₂16

    log₂ 16x⁴ + logb x²=log₂ 16

    logb x²=log₂ 16 - log₂ 16x⁴

    logb x²=log₂ (16/16x⁴)

    logb x² = log₂ x⁻⁴

    logb x²=log₂ x² / log₂ b

    log₂ x² / log₂ b=-log x⁴

    log₂ b=log₂ x² / - log₂ x⁴

    log₂ b=2 log x / - 4 log₂ x

    log₂ b=-1/2 ⇔ b=2⁻¹/² = 1/√2 = (√2) / 2≈ 0.707106781 ...

    Answer: b = (√2) / 2

    To chek

    if x=10

    log base √2/2 = log√2/2

    4 log₂ 20 + 2log√2/2 10=4 (4.321928095 ...) + 2 (-6.64385619=

    = 17.28771238-13.28771238=4
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