Logs

**M670** · December 8th 2012, 05:45 PM

OK so i have $\log_2(2x)+\log_2(x-3)=3$
I think I should rewrite it as follows $\log_2(2x(x-3))=3$ then what?
and I need help with $e^-2x+3\bullet e^-x=10$

**topsquark** · December 8th 2012, 05:49 PM

Originally Posted by M670

OK so i have $\log_2(2x)+\log_2(x-3)=3$
I think I should rewrite it as follows $\log_2(2x(x-3))=3$ then what?

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

-Dan

**topsquark** · December 8th 2012, 05:53 PM

Originally Posted by M670

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

First the LaTeX code. Instead of using \bullet I'd use \cdot. It looks a bit nicer.

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

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

So the equation is going to be:
$1 + 3 e^x = 10e^{2x}$

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

Now let $y = e^x$ . This gives
$10y^2 - 3y - 1 = 0$

Solve this then back substitute to get x.

-Dan

**M670** · December 9th 2012, 06:51 AM

Originally Posted by topsquark

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

-Dan

I have seen in the past that you can use this $2^{\log_2(a)} = a$ to get this $(2x\cdot(x-3))=3^2$ which I believe would give me $2x^2-6x-9=0$ 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?

**M670** · December 9th 2012, 07:05 AM

Originally Posted by M670

I have seen in the past that you can use this $2^{\log_2(a)} = a$ to get this $(2x\cdot(x-3))=3^2$ which I believe would give me $2x^2-6x-9=0$ 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= $\frac{6+\sqrt108}{4}$

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