1. ## Logs/Exponent Questions

A few question i cannot solve and need some help on:
1) $ln(5x)-ln(3-2x)=1$
2) $ln(x)+ln(3x+1)=1$
3) $8e^-x-e^x$

P.S

2. Originally Posted by Paymemoney
A few question i cannot solve and need some help on:
1) $log(5x)-log(3-2x)=1$
2) $log(x)+log(3x+1)=1$
3) $8e^-x-e^x$

P.S
I'll do the first as an example

$log(5x)-log(3-2x)=1$

$log\left(\frac{5x}{3-2x}\right) = 1$

$\frac{5x}{3-2x} = 10^1 = 10$

$5x = 30-20x$

$25x = 30$

$x = \frac{6}{5}$

3. ok, well I've solved 1) & 2) but how would i solve 3)?

4. Originally Posted by Paymemoney
ok, well I've solved 1) & 2) but how would i solve 3)?
To solve 3 it will need to be equal to something

As it stands you can only take out a factor of $e^{-x}$

$8e^{-x}-e^x = e^{-x}(8-e^{2x})$

5. i think the equation is meant to be $8e^{-x}-e^{x}=2$

6. Originally Posted by Paymemoney
i think the equation is meant to be $8e^{-x}-e^{x}=2$
In that case multiply throughout by $e^x$

$8-e^{2x}=2e^x$

Rearrange into the form $ax^2+bx+c=0$

$e^{2x}+2e^x-8=0$

Solve the quadratic (this one factorises) and then find x using the natural log. Remember that one solution will not be real.

I get solutions of

$x = ln(2)$ REAL

and

$x = ln(-4) = ln(4) + i\pi$ COMPLEX

You may only need the REAL solution

7. ok thanks