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
$\displaystyle 2e^{7x}+2=6$
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;
$\displaystyle x = \frac {\ln 2}{7}$
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 $\displaystyle e^{\ln(y)} = y$
$\displaystyle \ln (3x-4) + \ln (x-3)-\ln 4=0$
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.