1. ## log quation

Hi there, I have some simultaneous eqaution that are in the form of logarithm (1)$\displaystyle \log_{3}x = y= \log_{9}(2x-1)$ I started solving by removing he brackets from the secound equation

$\displaystyle 2x\log_{9}-\log_{9}$

$\displaystyle \frac{2x\log_{9}}{\log_{9}}$

then I got $\displaystyle y=2x$

from there I don't know if I am correct so far or what to do next...

I was going to put the other but I just think that I would give them a fair try first. can anyone please help me with this problem

2. Very bad! Logs can help you take apart MULTIPLICATION not addition. However, this does not seem to be what you have done, either. You seem to thing the "log" is some sort of factor. This is not the case. You cannot distribute a FUNCTION through to the elements of its argument.

"log" doesn't mean anything in this context. It MUST have an argument. "log(x)", now that means something.

In this problem, you are supposed to recognize and exploit this one thing, $\displaystyle 3^{2} = 9$. You will also use the base transformation rules.

$\displaystyle log_{3}(x) = log_{9}(2x-1)$

Transform to a common base.

$\displaystyle \frac{log(x)}{log(3)} = \frac{log(2x-1)}{log(9)}$

Now what?

3. Originally Posted by TKHunny
Very bad! Logs can help you take apart MULTIPLICATION not addition. However, this does not seem to be what you ahve done, either. You seem to thing the "log" is some sort of factor. This is not the case. You cannot distribute a FUNCTION through to the elements of its argument.

"log" doesn't mean anything in this context. It MUST have an argument. "log(x)", now that means something.

In this problem, you are supposed to recognize and exploit this one thing, $\displaystyle 3^{2} = 9$. You will also use the base transformation rules.

$\displaystyle log_{3}(x) = log_{9}(2x-1)$

Transform to a common base.

$\displaystyle \frac{log(x)}{log(3)} = \frac{log(2x-1)}{log(9)}$

Now what?
$\displaystyle \log x=\log 2(2x-1)$

$\displaystyle x=4x-2$

$\displaystyle -x3=-2$

$\displaystyle x=\frac {2}{3}$?

that can not be right the book has it as x=1

4. $\displaystyle \log_3(x) = \log_9(2x-1)$

$\displaystyle \log_3(x) = \frac{\log_3(2x-1)}{\log_3(9)}$

$\displaystyle \log_3(x) = \frac{\log_3(2x-1)}{2}$

$\displaystyle 2\log_3(x) = \log_3(2x-1)$

$\displaystyle \log_3(x^2) = \log_3(2x-1)$

$\displaystyle x^2 = 2x-1$

$\displaystyle x^2 - 2x + 1 = 0$

$\displaystyle (x-1)^2 = 0$

$\displaystyle x = 1$