[tex]\bf \cfrac{dx}{dt}=\cfrac{5}{t}\quad ,\quad x(0)=7\to
\begin{cases}
f(t)=0\\
t=7
\end{cases}\\\\
-----------------------------\\\\
\displaystyle \int \cfrac{5}{t}\cdot dt\implies 5\int \cfrac{1}{t}\cdot dt\implies 5ln|t|+C= f(t)
\\\\\\
\textit{now, we know when f(t)=0, t=7}\implies 5ln|7|+C=0
\\\\\\
C=-5ln(7)
\\\\\\
thus\implies \boxed{5ln|t|-5ln(7)=f(t)=x}[/tex]
notice |7| 7 is positive, so we can simply remove the bars and use 7 by itself
