College Algebra

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Solve the logarithmic equation. Express your solutions in exact form only. Please show all of your work.
log base 6 (64x^3+1)-log base 6 (4x+1)=1

  • College Algebra -

    log 6[ (64x^3+1)/(4x+1) ] = 1

    (64x^3+1)/(4x+1) = 6^1
    (4x+1)(16x^2 - 4x + 1)/(4x+1) = 6
    16x^2 - 4x - 5 = 0
    x= (4 ± √336)/32
    = (4 ± 4√21)/32
    = (1 ± 21)/8

    We have to reject the negative answer or else the log would be undefined
    so
    x = (1 + √21)/8

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