# Thread: Logarithmic problem baffling me...

1. ## Logarithmic problem baffling me...

Ok, I am fairly new to working with logarithms, but things had been going fairly well up until now. I had some minor mistakes here and there, and had gotten over the initial "hump" when you start something new, but now I am completely locked. The problem is:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) = -1

So, I add 1 to both sides, and then multiply it by (2 * 3^x + 3) / (2 * 3^x + 3) to get:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) + (2 * 3^x + 3) / (2 * 3^x + 3) = 0

I then add them together, and get:

(3^2x - 4 * 3^x + 3) / (2 * 3^x + 3) = 0

From here, I know that I can safely forget about the denominator, as it is only the numerator that is going to give me 0, so now I am at:

(3^2x - 4 * 3^x + 3) = 0

I put that through the quadratic formula, and find that 3^x is equal to 4, or 0. So I change this to a logarithm as such:

x log 3 = log 4

Divide both sides by log 3, and get log 4/log 3. Too bad the answer is 1 for this instance. (the other being 0, which at least works out) Can someone tell me where I have gone wrong? I have burnt 2 hours of my night trying to figure this out, and it is driving me absolutely crazy...

Thank you.

2. ## Re: Logarithmic problem baffling me...

$\displaystyle \frac{3^{2x} - 6 \cdot 3^x}{2 \cdot 3^x + 3} = -1$

let $\displaystyle t = 3^x$ ...

$\displaystyle \frac{t^2 - 6t}{2t+3} = -1$

$\displaystyle t^2 - 6t = -2t-3$

$\displaystyle t^2 - 4t + 3 = 0$

$\displaystyle (t - 3)(t - 1) = 0$

$\displaystyle t = 3 \implies 3^x = 3 \implies x = 1$

$\displaystyle t = 1 \implies 3^x = 1 \implies x = 0$

3. ## Re: Logarithmic problem baffling me...

Originally Posted by Latsabb
Ok, I am fairly new to working with logarithms, but things had been going fairly well up until now. I had some minor mistakes here and there, and had gotten over the initial "hump" when you start something new, but now I am completely locked. The problem is:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) = -1

So, I add 1 to both sides, and then multiply it by (2 * 3^x + 3) / (2 * 3^x + 3) to get:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) + (2 * 3^x + 3) / (2 * 3^x + 3) = 0

I then add them together, and get:

(3^2x - 4 * 3^x + 3) / (2 * 3^x + 3) = 0

From here, I know that I can safely forget about the denominator, as it is only the numerator that is going to give me 0, so now I am at:

(3^2x - 4 * 3^x + 3) = 0

I put that through the quadratic formula, and find that 3^x is equal to 4, or 0. So I change this to a logarithm as such:
That is incorrect.
$\displaystyle 3^x=3\text{ or }1$

4. ## Re: Logarithmic problem baffling me...

Originally Posted by Latsabb
Ok, I am fairly new to working with logarithms, but things had been going fairly well up until now. I had some minor mistakes here and there, and had gotten over the initial "hump" when you start something new, but now I am completely locked. The problem is:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) = -1

So, I add 1 to both sides, and then multiply it by (2 * 3^x + 3) / (2 * 3^x + 3) to get:

(3^2x - 6 * 3^x) / (2 * 3^x + 3) + (2 * 3^x + 3) / (2 * 3^x + 3) = 0
A somewhat simpler thing to do would be to multiply both sides by the denominator 2*3^x+ 3 to get 3^(2x)- 6(3^x)= -2(3^x)- 3 which we can then
write as (3^x)^2- 4(3^x)+ 3= 0, a quadratic equation in 3^x. Letting y= 3^x, we have y^2- 4y+ 3= (y- 3)(y- 1)= 0 so that y= 3^x= 3 and y= 3^x= 1, not "4 and 1".

I then add them together, and get:

(3^2x - 4 * 3^x + 3) / (2 * 3^x + 3) = 0

From here, I know that I can safely forget about the denominator, as it is only the numerator that is going to give me 0, so now I am at:

(3^2x - 4 * 3^x + 3) = 0

I put that through the quadratic formula, and find that 3^x is equal to 4, or 0. So I change this to a logarithm as such:

x log 3 = log 4

Divide both sides by log 3, and get log 4/log 3. Too bad the answer is 1 for this instance. (the other being 0, which at least works out) Can someone tell me where I have gone wrong? I have burnt 2 hours of my night trying to figure this out, and it is driving me absolutely crazy...

Thank you.

5. ## Re: Logarithmic problem baffling me...

Oh my god... I see where I went wrong. I was seriously confused when I looked at what you guys posted, and everyone had the same things as me all the way until the quadratic equation. I went back, and looked at my papers, and see my problem. In the squareroot, I had 16-12, which is correct, but for some reason I didnt root it, so I had (4+-4)/2. I seriously dont see how I overlooked this so many times last night... I was so focused on it being a problem somewhere else, that I didnt look hard enough at the quadratic section I guess...

I seriously hate small errors that cost so much time! But thank you all. I had a bad nights sleep over the problem, lol.