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Jordan
Lv 7
Jordan asked in Science & MathematicsMathematics · 8 years ago

Solve an equation containing a log?

How do you solve this equation?

log(3x+1)=2

Update:

That is how the equation is listed, so I would say its safe to assume its log base 10.

6 Answers

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  • Raymond
    Lv 7
    8 years ago
    Favorite Answer

    Find the "antilog"

    If the log is in base 10, then the "antilog" of 2 is 10^2 (equal to 100)

    the antilog of a log is simply the "argument" of the log.

    (if the base is 10)

    10^(log(3x+1)) = 3x+1

    Therefore, by taking the antilog on both sides, you would get

    3x + 1 = 10^2

    (if the base is not 10, then simply use the correct base):

    3x + 1 = (base)^2

  • Ed
    8 years ago

    Assuming you mean log to the base of 10,

    3x + 1 = 10^2 = 100 so 3x = 99 so x = 33

  • lenpol7
    Lv 7
    8 years ago

    log(3x+1)=2

    Antilog

    3x + 1 = 10^2

    3x + 1 = 100

    3x = 99

    x = 33

  • Professor Of Wumbology
    8 years ago

    log(b) = a just means 10^a = b.

    "log(3x+1) = 2" turns into "10^2 = 3x+1"

    10^2 = 3x + 1

    100 = 3x + 1

    99 = 3x

    x = 33

    Source(s): My cone-shaped head.
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  • Chris
    Lv 4
    8 years ago

    antilog 2 = 100

    Hence log 100 = 2

    Hence 3x + 1 = 100

    Hence 100 - 1 = 99 = 3x

    Therefore x = 99/3 = 33

  • Rita the dog
    Lv 7
    8 years ago

    3x+1 = 10^2, so 3x=99 and x=33.

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