# How do you solve for y in Ln(y - 7) - ln(7) = x + ln(x)?

Mar 28, 2016

$y = y = {e}^{x} 7 x + 7$

#### Explanation:

$1$. Start by using the natural logarithmic property, ${\ln}_{\textcolor{p u r p \le}{b}} \left(\frac{\textcolor{red}{m}}{\textcolor{b l u e}{n}}\right) = {\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) - {\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the left side of the equation.

$\ln \left(y - 7\right) - \ln \left(7\right) = x + \ln \left(x\right)$

$\ln \left(\frac{y - 7}{7}\right) = x + \ln \left(x\right)$

$2$. Use the natural logarithmic property, ${\ln}_{\textcolor{p u r p \le}{b}} \left({\textcolor{p u r p \le}{b}}^{\textcolor{\mathmr{and} a n \ge}{x}}\right) = \textcolor{\mathmr{and} a n \ge}{x}$, to rewrite $x$ on the right side of the equation.

$\ln \left(\frac{y - 7}{7}\right) = \ln \left({e}^{x}\right) + \ln \left(x\right)$

$3$. Use the natural logarithmic property, ${\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m} \cdot \textcolor{b l u e}{n}\right) = {\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) + {\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the right side of the equation.

ln((y-7)/7)=ln(e^xx)

$4$. Since the equation now follows a "$\ln = \ln$" situation, where the bases are the same on both sides, rewrite the equation without the "$\ln$" portion.

$\frac{y - 7}{7} = {e}^{x} x$

$5$. Solve for $y$.

$y - 7 = {e}^{x} 7 x$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = {e}^{x} 7 x + 7 \textcolor{w h i t e}{\frac{a}{a}} |}}}$