# Thread: Another Logarithm Equation: Not sure how to solve for x properly.

1. ## Another Logarithm Equation: Not sure how to solve for x properly.

Hi again,

I am stuck on another logarithmic equation:

log3 x - 2 = log3 (x - 8)

Now, I assume log3 cancels out, which leaves:

x - 2 = x - 8

But from there, I am unsure how you would get x by itself, because if you try and transfer x to just one side, it cancels out because you have x and -x. I think I am on the wrong track though, so all guidance will be much appreciated once again.

Thanks,

Nathaniel

2. Originally Posted by BinaryBoy
Hi again,

I am stuck on another logarithmic equation:

log3 x - 2 = log3 (x - 8)

Now, I assume log3 cancels out, which leaves:

x - 2 = x - 8 Mr F says: It should be clear that this is nonsense as it implies -2 = -8.

But from there, I am unsure how you would get x by itself, because if you try and transfer x to just one side, it cancels out because you have x and -x. I think I am on the wrong track though, so all guidance will be much appreciated once again.

Thanks,

Nathaniel
The equation is possibly meant to be $\displaystyle \log_3 (x) - 2 = \log_3 (x - 8) \Rightarrow \log_3 \left( \frac{x}{x - 8} \right) = 2$ (and your job is to fill in the missing steps in getting this).

Now solve for x.

3. If the problem was, indeed, $ln_3(x- 2)= ln_3(x- 8)$, then what you did is exactly correct- ln(x) is a "one-to-one function" so it must be true that x- 2= x- 8. But no matter what x is, that is the same as -2= -8 which is not true. NO value of x makes that equation true.

But,as mr fantastic said, your lack of parentheses makes this problem ambiguous. It might be $ln_3(x- 2)= ln_3(x- 8)$, $ln_3(x)- 2= ln_3(x- 8)$, $ln_3(x- 2)= ln_3(x)- 8$, or $ln_3(x)- 2= ln_3(x)- 8$. The first and last of those, which would be the most reasonable interpretations, have no solution!

4. Thanks for the help Mr Fantastic and HallsofIvy. I have solved for x and checked my answer and the equation is indeed of the form: log3 (x) -2 = logx (x - 8).

So, what I did was:

log3(x) = logx (x - 8) + 2

Then I moved the logx(x - 8) to the other side by dividing both sides. So I get:

log3(x / x - 8) = 2

This turns into:

log3( x / x - 8) = (3 to the power of 2)
= 9

Then I multiplied the x - 8 back over:

log3(x) = 9(x - 8)

log3(x) = 9x - 72

Multiply the x over as well:

0 = 8x - 72

Move the 72 to the other side

8x = 72

Divide by 8 to yield:

x = 9.

Cool! Many thanks for the help!

Nathaniel

5. You should, of course, check that answer in the original problem:
$log_3(x)- 2= log_3(x- 8)$

Setting x= 9, $log_3(9)= log_3(3^2)= 2$ and $log_3(9- 8)= log_3(1)= log_3(3^0)= 0$ so the equation becomes 2- 2= 0 which is, in fact, correct.