solve p(1+q)=qz (partial differential equation)

Question

this is from partial differential equation chapter and final answer is logaz-1=x+aylogbe

Answer 1

Could you explain your answer to me, the final one: logaz-1=x+aylogbe

Is the left hand side supposed to be,

log(az-1)

log(az) - 1

log_a(z) - 1     (the log of z, to the base a, minus 1)

or something else ?

Is the right hand side supposed to be

ay*log(be)

ay*log_b(e)    (ay times the log of e, to the base b)

or something else ?

Are a, b, e constants?

is e the base of natural logarithms?

Does log mean natural logarithm or common logarithm ?

Answer 2

p(1+q)=qz        .......1

which is the form of(p,q,z)

p=dz/dx=dz/du*du/dx

p=dz/du*1        (du/dx=1)

q=dz/du*du/dy

q=dz/du*a      (du/dy=a)

putt the value of p and q ......1

dz/du(1+adz/du)=azdz/du

(1+adz/du)=az

adz/du=az-1

Answer 3

Let z be a function of u such that ,u=x+ay. Then, p=dz/du and q=a*dz/du. Substituting this in the equation, dz/du(1+a*dz/du)=za*dz/du . By simplification we will get , (a/(az-1))dz=du. Therefore answer is log(az-1)=x+ay+C