Solve for x in this problem

If c does not equal 1, e^ax=ce^bx I have no idea how to solve this

e^(ax) = c*e^(bx)
ln[e^(ax)] = ln[c * e^(bx)]
ln[e^(ax)] = ln(c) + ln[e^(bx)]
ax - bx = ln(c)
x(a - b) = ln(c)
x = ln(c)/(a - b)

e^ax = ce^bx
e^ax = c•e^bx, where '•' plays a multiplicative function, then apply the natural logarithmic function 'In' to both sides, and you'll obtain
ln e^ax = ln (c•e^bx)
=> In natural logarithm, In (a•b) = In a + In b, then by applying this rule at the right hand side, you'll obtain
ln e^ax = ln c + ln e^bx
=> Consider this rule: In e^a = a, then
ax = ln c + bx
ax - bx = In c
x(a - b) = ln c
x = ln c/(a - b) ...Ans.

The only way you can do this type is to take Ln of each side, getting ax = Ln(c) + bx; that gives us ax - bx = Ln(c) ; take out a common factor of x and x(a - b) = Ln(c) so divide by (a - b) and x = Ln(c)/(a-b).

Consider taking the natural logarithm of both sides and remember that the logarithm of a product is the sum of logarithms for each factor.

ANS: x = c/(a-b)

work:
e^ax=ce^bx = e^cbx
ax = cbx or
x = c/(a-b)

axlne=bx(lnc+lne)
note:lne=1
ax-bxlnc=1
x(a-blnc)=1
x=1/(a-blnc)