Solving log equations?

Could someone help me on this problem? I have a math exam tomorrow and I'm having trouble finding examples online:

log(3x + 7) + log(x - 2) = 1

Thank you in advance!

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  • 1 decade ago
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    Assuming you are using the base 10 logarithm, if we take 10th power of both sides of the equality in question, then we have

    10^(log(3x+7)+log(x-2)) = 10

    using properties of logarithms which I'm sure you know, we simplify the left side and the problem is turned into solving the quadratic equation

    (3x+7)(x-2) = 10 , which have 2 solutions: x=8/3, x=-3

    but obviously, when we took the exponential, we've introduced some new properties into the equality that wasn't there originally, therefore we must check the solutions we have for x in the original equality

    x=8/3 does not violate any properties, but x=-3 turns the argument in the second logarithm negative, which we cannot have.

    \therefore, x= 8/3 is the only solution for the equality in question

  • Anonymous
    1 decade ago

    Remember log (a*b)=log a + log b ?

    Use it in reverse:

    log(3x + 7) + log(x - 2) =log[(3x + 7) (x - 2)]

    and 1 is just log 10

    so you have log[(3x + 7) (x - 2)]=log 10

    log[(3x + 7) (x - 2)}-log 10=0

    Remember log a/b=log a - log b?

    log[(3x + 7) (x - 2)/10]=0

    But 0=log 1, so the expression in the [ ] must be =1

    (3x + 7) (x - 2)/10=1

    Simplify

    3x^2 -6x+7x-14=10

    3x^2+x-24=0

    Solve it.

    There are 2 solutions. Within the realm of real numbers, only one of them is acceptable, the positive one.

    While actually both are acceptable, you have to use complex numbers, which I don't think you've seen, yet.

  • Anonymous
    1 decade ago

    log(3x + 7) + log(x - 2) = 1

    so according to log rules

    log((3x + 7)*(x - 2)) = 1

    10^(log((3x + 7)*(x - 2)) = 10^1

    (3x + 7)(x - 2) = 10

    3x^2 + x - 14 = 10

    3x^2 + x - 24 = 0

    x = [-1 +/- sqrt(1 - 4(3)(-24))]/(2*3)

    x = [-1 +/- sqrt(289)] / 6

    x = -3, 2.667

    -3 gets thrown out because you can't take a log of a negative number, so 2.666... is the answer.

    If you need help understanding why this works, let me know and I'll edit it in.

  • 1 decade ago

    log(3x^2+x-14)=log10

    3x^2+x-24=0

    x=[-1+ & -(1+288)^1/2]/6=8/3 & -3

    only x=8/3 is acceptable

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  • 1 decade ago

    logA +logB=log(A.B)

    Apply this property in the eqn. to get

    log(3x.x+x-14)=1

    Use definition of log( if logx to base a = b, then a to the power b=x)

    10 power 1 =3x.x+x-14

    => 3x.x+x-24=0

    x=-3,8/3

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