Do we reduce the problem into the known $3^x+4^x=5^x $? I want to solve the following equations: \begin{align*}&(i) \ \ \ \ \ 3^x+4^x=5^x \\ &(ii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+3)^x, \ x>-1 \\ &(iii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+4)^x, \ x>-1\end{align*} 
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I have done the following: 
For (i): 
We have that $$3^x+4^x=5^x\Rightarrow \left (\frac{3}{5}\right )^x+\left (\frac{4}{5}\right )^x=1$$ This equation of the form $f(x)=a^x+b^x$ with $0<a<b<1$. For this equation we have that: Since $0<a,b<1$ the function $a^x$ and $b^x$ and so also $f(x)$ are decreasing. 
So there is at most one solution. 
Since $a<b$ so $a^x<b^x$ for positive $x$ then we get $2a^x<f(x)<2b^x$. Both the lower and upper bound for $f(x)$ pass through 1, we know that f(x) must also pass through 1, so there is exactly one solution,. 
We symbolize by $c$ the unique solution. From $2a^c<1<2b^c$ we get the inequality $-\log_a2<c<-\log_b2$. For $a=\frac{3}{5}$ and $b=\frac{4}{5}$ we get $1.35<c<3.10$. We consider the natural numers between $1.35$ and $3,10$. 
We see that for $c=2$ the equation is satisfied, and so $x=2$ is the only solution. 
Is everything correct? 
As for (ii) and (iii) do we have to find an appropriate $x$ so that we can reduce it to a problem as in (i) ? 

Another approach
$x=2$ is an obvious solution to the problem. So consider the function
$$f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1.$$
We know $f(2)=0$. Also $f'(x)=\left(\frac{3}{5}\right)^x\ln\left(\frac{3}{5}\right)+\left(\frac{4}{5}\right)^x \ln\left(\frac{4}{5}\right) <0$ for all $x$. Should this $f$ had another zero, say $f(c)=0$, then by Rolle's theorem, there exists a point $k \in (c,2)$ (or $(2,c)$) such that $f'(k)=0$ but thay contradicts the sign of the derivative we have shown above. Thus only one solution.  
