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

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