Solving harder exponential equations, e.g. 5/[exp(x) + 6exp(-x)] - 1 = 0 . CORE MATHS.

Our example equation is, 5/[exp(x) + 6exp(-x)] - 1 = 0 ,and we wish to solve for x. Note that exp(x) is the same as writing e^x I have chosen this one, because it helps illustrate some general good practice in solving equations. The first step I would reccommend, would be to multiply the equation through by the denominator in the first term on the left-hand side. Generally, denominators (especially long ones) aren't desirable. We get,5 - [exp(x) + 6exp(-x)] = 0 . Here, it is usually helpful to expand the brackets. So, 5 - exp(x) - 6exp(-x) = 0 . Here is where the tip isn't so general. When solving exponential equations like this, it is usually helpful to introduce a new variable y = exp(x), to make the solution clearer. So,5 - y - 6(1/y) = 0Again, we should avoid denominators. So multiplying through by y gives us, 5y - y² - 6 = 0 . This should be looking very familiar to most. It is a quadratic equation. Rearrangement is optional at this point, but I prefer to write this equation as.y² - 5y + 6 = 0 . Simply factorising,(y - 2)(y - 3) = 0Hence,y = 2 OR y = 3 . But let's not forget, the aim was to find x. Remembering that y = exp(x), we have. exp(x) = 2 OR exp(x) = 3 . To get x, we need to find the numbers such that when we raise e to the power of them, we get either 2 or 3. Luckily, we have a function to do just that. Our final answer is, x = ln(2) OR x = ln(3) , where ln is the natural logarithm (log base e).