1. ## Logs

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 $\displaystyle e^-2x+3\bullet e^-x=10$

2. ## Re: Logs

Originally Posted by M670
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?
So far so good. Now, since $\displaystyle 2^{\log_2(a)} = a$ what's your next step?

-Dan

3. ## Re: Logs

Originally Posted by M670
and I need help with $\displaystyle e^-2x+3\bullet e^-x=10$
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

4. ## Re: Logs

Originally Posted by topsquark
So far so good. Now, since $\displaystyle 2^{\log_2(a)} = a$ what's your next step?

-Dan
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?

5. ## Re: Logs

Originally Posted by M670
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?
So I believe my answer is x= $\displaystyle \frac{6+\sqrt108}{4}$