Number of solutions of the equation $3^x+4^x=5^x$ in the set of positive real numbers [duplicate]

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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?

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

• 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. – abiessu Jun 5 '15 at 6:50
• @abiessu yes, I tried putting positive integers like x=1, x=2 etc. But I wanted to know how can it be solved algebraically... – Ritu Jun 5 '15 at 6:52
• The simple answer is "it can't". – abiessu Jun 5 '15 at 6:53

Divide both sides by $5^x$ to get $\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x = 1$.
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.
• @Ritu Are you sure that $3^1 + 4^1 = 5^1$? – MathMajor Jun 5 '15 at 6:54
• @Ritu: can you show the math for $x=1$? – abiessu Jun 5 '15 at 6:54
• 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. – JimmyK4542 Jun 5 '15 at 6:56