NMTC Practice Part 12

divide both sides by (2014)^x. define f(x) = (2015/2014)^x - (2013/2014)^x - 1 this is continuous and differentiable. f'(x) = ln(2015/2014) * (2015/2014)^x + ln(2014/2013) * (2013/2014)^x >0 thus f(x) is strictly increasing. f(0) = -1 also as x tends to infinity, f(x) tends to infinity and thus there must be exactly one value where f(x) is 0

My solution is similar to Hemang's. Dividing both sides by $2015^x$ , we can write the equation as $(2013/2015)^x+(2014/2015)^x=1$ . The LHS, being the sum of two functions of exponential decay, is decreasing with a range of $(0,\infty)$ . Thus our equation has exactly $\boxed{1}$ solution .

Divide both sides of equation by (2015^x). It becomes: (2013/2015)^x + (2014/2015)^x = 1.

Since the left hand side can take all possible positive values when x ranges over the real numbers, there is at least one solution for x. On the other hand, since the left hand side is strictly decreasing function, there is at most one solution for x. Thus there is exactly one solution for x, this solution can be found by differentiating the above equation and solving

2013^x+2014^x=2015^x Simplify the left, then you get 4027^x=2015^x. X=0