Hi, I have the following equations that I am really confused on..just wondering if someone can help me out with them..thanks!

1) 2e^7x+2=6

2) 1n (3x-4) + 1n (x-3)-1n 4=0

2. $\displaystyle 2e^{7x}+2=6$

Subtract 2 from both sides:

$\displaystyle 2e^{7x}=4$

Divide both sides by 2:

$\displaystyle e^{7x}=2$

Take the natural log of both sides:

$\displaystyle \ln e^{7x}=\ln 2$

Recall this law of logarithms:

$\displaystyle log (x)^a = a\cdot log (x)$

Apply it:

$\displaystyle 7x\cdot \ln e = \ln 2$

The natural log of e is 1, so we have:

$\displaystyle 7x = \ln 2$

Divide both sides by 7;

$\displaystyle x = \frac {\ln 2}{7}$

3. For problem 1 rearrange your equation so that your exponentiated factor ($\displaystyle e^{7x}$) is by itself (no coefficient) then treat each side of the equation like a variable and plug each into the inverse of $\displaystyle e^y$. (hint: You should be using the fact that $\displaystyle \ln(e^y) = y$.)

For problem 2 observe that $\displaystyle \ln(a)+\ln(b) = \ln(ab)$ and as an immediate consequence $\displaystyle \ln(c) - \ln(d) = \ln(c/d)$. Also note that $\displaystyle e^{\ln(y)} = y$

4. $\displaystyle \ln (3x-4) + \ln (x-3)-\ln 4=0$

Recall this law of logarithms:

$\displaystyle \log (a) + \log (b) = \log (ab)$

Apply it:

$\displaystyle \ln \left((3x-4)\cdot (x-3)\right) - \ln 4 = 0$

$\displaystyle \ln (3x^2 - 13x + 12) - \ln 4 = 0$

$\displaystyle \ln (3x^2 - 13x + 12) = \ln 4$

$\displaystyle e^{\ln 4} = 3x^2 - 13x + 12$

Recall that:

$\displaystyle e^{\ln x} = x$

Apply it:

$\displaystyle 3x^2 - 13x + 12 = 4$

$\displaystyle 3x^2 - 13x + 8 = 0$

Solve for x.