2
$\begingroup$

This question already has an answer here:

I came across the following question:

Find the number of solutions of the equation $3^x+4^x=5^x$ in the set of positive real numbers.

I tried the above question by taking log on both sides and then solving, but it didn't seem to work. Any idea on how to proceed to solve this question?

$\endgroup$

marked as duplicate by Travis Willse, Community Jun 5 '15 at 7:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Did you already account for the solution $x=2$? Proceeding from there, divide though by $5^x$ and note that the LHS is strictly decreasing over the positive reals, showing that there can be only one solution. $\endgroup$ – abiessu Jun 5 '15 at 6:50
  • $\begingroup$ @abiessu yes, I tried putting positive integers like x=1, x=2 etc. But I wanted to know how can it be solved algebraically... $\endgroup$ – Ritu Jun 5 '15 at 6:52
  • $\begingroup$ The simple answer is "it can't". $\endgroup$ – abiessu Jun 5 '15 at 6:53
6
$\begingroup$

Divide both sides by $5^x$ to get $\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x = 1$.

The left side is strictly decreasing, while the right side is constant. Hence, there is at most one solution. All you have to do is find one solution, and you have found all solutions.

EDIT: As far as solving this algebraically goes, for most values of $a,b,c > 0$, the equation $a^x+b^x = c^x$ doesn't have a nice solution for $x$ in terms of $a,b,c$. However, for specific values of $a,b,c$ there can be a nice solution. If this is a textbook problem for which an exact answer is required, then the authors will have picked out values of $a,b,c$ for which there is a nice solution.

$\endgroup$
  • $\begingroup$ but the equation is true for two values so far like x=1, x=2.... $\endgroup$ – Ritu Jun 5 '15 at 6:53
  • $\begingroup$ @Ritu Are you sure that $3^1 + 4^1 = 5^1$? $\endgroup$ – MathMajor Jun 5 '15 at 6:54
  • $\begingroup$ @Ritu: can you show the math for $x=1$? $\endgroup$ – abiessu Jun 5 '15 at 6:54
  • $\begingroup$ oky, i got it.. my hand and mind didn't syncronise... $\endgroup$ – Ritu Jun 5 '15 at 6:55
  • 1
    $\begingroup$ Actually, no, the equation is not true for $x = 1$ since $3^1+4^1 = 7 \neq 5 = 5^1$. However, it is true for $x = 2$, as you have found out. $\endgroup$ – JimmyK4542 Jun 5 '15 at 6:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.