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Today, 01:44

Solve for x

27=9^-x+5

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Answers (2)
  1. Today, 03:49
    0
    27 = 9^-x + 5

    Subtract 5 from both sides

    22 = 9^-x

    Log 22 = - x log 9

    (Log22) / (log9) = - x

    - (log22) / (log9) = x

    -1.41 = x
  2. Today, 04:43
    0
    1. Use negative power rule: x^-a=1/x^a

    27=1/9^x+5

    2. Subtract 5 from both sides

    27-5=1/9^x

    3. Simplify 27-5 to 22

    22=1/9^x

    4. Multiply both sides by 9^x

    22*9^x=1

    5. Divide both sides by 22

    9^x=1/22

    6. Use definition of common logarithm: b^a=x if and only if log (x) = a

    x=log9 (1/22)

    7. Use change of base rule: logbx = (logan) / 8logab)

    x = (log1/22) / (log9)

    8. Use power rule: logbx^c=clogbx log1/22 - > log (22^-1) - > - log22

    x = (-log22) / (log9)

    9. Use power rule

    Use power rule: logbx^c=clogbx, log9 - > log3^2 - > 2 log3

    x = (-log22) / (2log3)

    10. Your answer is, and my wish of you having a nice day : D

    x = - (log22) / (2log3)

    x = (approx.) - 1.41
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