OK so i have $\displaystyle \log_2(2x)+\log_2(x-3)=3$

I think I should rewrite it as follows $\displaystyle \log_2(2x(x-3))=3$ then what?

and I need help with $\displaystyle e^-2x+3\bullet e^-x=10$

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- Dec 8th 2012, 05:45 PMM670Logs
OK so i have $\displaystyle \log_2(2x)+\log_2(x-3)=3$

I think I should rewrite it as follows $\displaystyle \log_2(2x(x-3))=3$ then what?

and I need help with $\displaystyle e^-2x+3\bullet e^-x=10$ - Dec 8th 2012, 05:49 PMtopsquarkRe: Logs
- Dec 8th 2012, 05:53 PMtopsquarkRe: Logs
First the LaTeX code. Instead of using \bullet I'd use \cdot. It looks a bit nicer.

$\displaystyle e^{-2x} + 3 \cdot e^{-x} = 10$

For the sake of clarity, I'd multiply both sides by $\displaystyle e^{2x}$. This step is not vital and you can easily alter the substitution I'm going to use.

So the equation is going to be:

$\displaystyle 1 + 3 e^x = 10e^{2x}$

$\displaystyle 10e^{2x} - 3e^x - 1 = 0$

Now let $\displaystyle y = e^x$. This gives

$\displaystyle 10y^2 - 3y - 1 = 0$

Solve this then back substitute to get x.

-Dan - Dec 9th 2012, 06:51 AMM670Re: Logs
I have seen in the past that you can use this $\displaystyle 2^{\log_2(a)} = a$ to get this $\displaystyle (2x\cdot(x-3))=3^2$ which I believe would give me $\displaystyle 2x^2-6x-9=0$ then I solve it using the quadratic formula. But I don't know the reason why I am able to square it and then drop the log off?

- Dec 9th 2012, 07:05 AMM670Re: Logs