What is the solution to: log (3x+9) = 1+log(x-3)? Please show all steps. My neice got it wrong on her test.?

Question

and she's not understanding why. I want to show her the correct way - but I do not know a thing about Logarithms. (sp?)

You both get thumbs up for answering, but I'm thinking Dave has the correct answer. We'll verify with her professor and then pick best answer! Thanks guys!

Answer 1

You can take 10 to the power of both sides, and you still have a valid equation:

10^(log(3x+9)) = 10^(1 + log(x-3))
10^(log(3x+9)) = 10^1 * 10^(log(x-3))

This is one identity that you need to know:

10^(log(x)) = x

So, moving forward:

3x + 9 = 10 * (x - 3)
3x + 9 = 10x - 30
7x = 39
x = 39/7

Answer 2

I'll work the problem in natural logs and base 10 logs.
Rewrite the equation as
log(3x+9) - log(x-3) = 1.
The difference of 2 logs is the log of their quotient.
So we get
log( (3x+9)/(x-3)) = 1
(3x+9)/(x-3) = e.
3x+9 = e(x-3)
x(3-e) = -9-3e
x = -9-3e/3-e.
But if we compute x, its approximate value is about
-60.89. And then 3x+9 is negative. Since we
cannot take the logarithm of a negative number,
there is no solution to the problem in natural logs.
In base 10, we get
3x + 9 = 10(x-3)
7x = 39
x = 39/7.
Now let's check our work for base 10 logarithms:
3x + 9 = 117/7 + 63/7 = 180/7
x-3 = 18/7
180/7 / 18/7 = 10
log 10 = 1, so done!

Answer 3

log (3x +9) = 1 + log(x - 3)
log(3x +9) - log(x -3) = 1
we have ,log 10 = 1 , and log (x / y) = log x - log y , so
log[( 3x + 9) / (x - 3)] = log 10
(3x +9 ) / (x - 3) = 10
3x + 9 = 10(x - 3)
3x + 9 = 10x - 30
7x = 39 , so
x = 39 / 7
solution set is { 39 / 7}

Answer 4

log(3x+9)=1+log(x-3)
this could be written as
log(3x+9)-log(x-3)=1
log(a)-log(b)=log(a/b)
log[ (3x+9)/(x-3)] = 1
(3x+9)(x-3)=10 ( base 10: it is 10 to the power 1)
(3x^2-9x+9x-27)=10
3x^2-27=10
3x^2=37
x^2=37/3
x=+/-sqrt(37/3)