Solving a differential equation $\int_{0}^x k/x \,\, dx = \int_{0}^t dt$ and $k\,\,dx/dt = x$, where $x=x(t)$ and $x(0) = 0$. In solving a differential equation $\int_{0}^x k/x\,\, dx = \int_{0}^t dt$ where I tried following:
$$\int_{0}^x \frac{k}{x} dx = \int_{0}^t dt$$
$$k[\ln x]_0^x = t$$
where $k$ is constant.
But $\ln 0$ is $-\infty$, so this can't be right. But Wolfram alpha does produce a result. What did I go wrong?
Edit: The original differential equation preceding above is $k\,\,dx/dt = x$, where $x=x(t)$ and $x(0) = 0$. Is it different from the above equation?
 A: Since you tagged "differential equations", I suppose that you had at a time $$k\frac{dx}{x}=dt$$ Integrating both sides, you then have $$k\log(x)=t+c$$ that is to say $$x=c e^{t/k}$$ and $c$ is defined by the initial condition given for $t=0$. So $$x=x_0~e^{t/k}$$ 
In others words, finish the integration work before speaking about bounds which do not have room here.
I hope and wish this is clarifying things to you. If not, please post.
A: The differential equation $k\frac{dx}{dt}=x$ has a simple rate of change interpretation. If at $t=0$ $x=0$, then the rate of change of $x$ at $t=0$ is $0$. Taking the time derivative of the differential equation gives $k\frac{d^2x}{dt^2}=\frac{dx}{dt}=\frac{x}{k}$ and again we find the derivative equals $0$. All higher derivatives $=0$ by the same ideas. So $x=0$ for all subsequent time.
In a sense the question is bad, because the integral $\int_0^x\frac{dx}{x}$ is really the integral $\int_0^0\frac{dx}{x}$. Something which grows in proportion to itself cannot suddenly come into existence.
