Solve the equation 5^(2x) - 12(5^x) + 35 = 0

The first step to solving this is equation is to notice that the equation is of a similar to the form of a quadratic equation: ay^2 + by + c  = 0 where a, b and c are constants. Next we introduce a new variable z = 5^x to reduce the equation to z^2 - 12 z + 35 = 0. Now we just need to solve the quadratic equation in z. First, we check if this is factorisable i.e. can be written as (z+a)(z+b) = 0. So we compare z^2 - 12 z + 35 and (z + a)(z + b) = z^2 + (a+b) z + ab. Therefore ab = 35 and a+b = -12. Next lets note that the factors of 35 are 1, 5, 7 and 35 as a and b have to be factors of 35. By considering the factors it should be clear that the choice of a is -5 aand b is -7 (note it does not matter which of -5 and -7 we assign to a and b). 

So we can write the quadratic equation in z as (z-5)(z-7) = 0 the solutions are z = 5 because this makes the first bracket = 0 and z = 7 because this makes the second bracket = 0. We defined z at the beginning as z = 5^x. Therefore, 5^x = 5 and 5^x = 7. The solution to the former of these equations is obviously 1 as 5^1 = 5. However, the solution to the second eqaution is not as obvious, so we take the natural log of both sides of the equation (any log could be taken as long as the base of the log was the same). So we get ln(5^x) = ln(7). Using the rule ln(a^x) = x ln(a) the equation reduces to x ln(5) = ln(7) hence x = ln(7)/ln(5). So the solution to the equation is x = 1 or x = ln(7)/ln(5)