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# 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!

### 5 Answers

- 1 decade agoFavorite Answer
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

- Anonymous1 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.

- Anonymous1 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.

- nozar nazariLv 71 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