1. ## Logs/Exponent Questions

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

P.S

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

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

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

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

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

$\displaystyle 5x = 30-20x$

$\displaystyle 25x = 30$

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

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 $\displaystyle e^{-x}$

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

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

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

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

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

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

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

$\displaystyle x = ln(2)$ REAL

and

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

You may only need the REAL solution

7. ok thanks