I've solved the DE:

$$a\cdot x(t)+b\cdot\ln\left(1+c\cdot x(t)\right)=-p\cdot x'(t)\tag1$$

Where $x(0)=k$

And I got the following solution:

$$t=\int_{x(t)}^k\frac{p}{a\cdot z+b\cdot\ln\left(1+c\cdot z\right)}\space\text{d}z\tag2$$

If I want to plot the solution in mathematica I get on the y-axis the time $t$ but I want time at the x-axis, how can I solve that?

I've solved the DE:

$$a\cdot x(t)+b\cdot\ln\left(1+c\cdot x(t)\right)=-p\cdot x'(t)\tag1$$

Where $x(0)=k$

And I got the following solution:

$$t=\int_{x(t)}^k\frac{p}{a\cdot z+b\cdot\ln\left(1+c\cdot z\right)}\space\text{d}z\tag2$$

If I want to plot the solution in mathematica I get on the y-axis the time $t$ but I want time at the x-axis, how can I solve that?

The code is equal to:

Plot[Integrate[(5*10^(-3))/((78/10)*
      z + ((((5463/
              20) + (20))*(((138064852)/(100000000))*10^(-23))*(2))/((\
(16021766208)/(10000000000))*10^(-19)))*Log[1 + ((z)/(10^(-4)))]), {z,
    x, Pi}], {x, 0, 1}]

Only the axis needs to be swaped.