How do you solve

I would like to lớn prove sầu that the equation $ 3^x+4^x=5^x $ has only one real solution ($x=2$)

I tried khổng lồ study the function $ f(x)=5^x-4^x-3^x $ (in order lớn use the intermediate value theorem) but I am not able lớn find the sign of $ f'(x)= ln(5) imes5^x-ln(4) imes4^x-ln(3) imes3^x $ and I can"t see any other method khổng lồ solve sầu this exercise...




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One direct method is lớn divide directly by $5^x$ and get $1=(3/5)^x+(4/5)^x$. From here it is clear that the RHS is strictly decreasing, and there is a unique solution. Almost all exponential equations can be treated this way, by transforming them to

one increasing function equal to one decreasing function

one increasing/decreasing function equal khổng lồ a constant.


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If we insert the known solution we can write $$ 5^2+x = 4^2+x + 3^2+x $$asking, whether there might another solution exist besides $x=0 $ . Then we can rewrite, putting the $5^x$ lớn the rhs:$$ 5^2 = 4^2cdot 0.8^x + 3^2cdot 0.6^x $$Then if the exponents $x$ on the rhs are zero, we have sầu the known solution. But if $x$ increases over zero, then the values of both summands decrease simultaneously, thus the eunique can no more hold.The analogue occurs for decreasing $x$: both summands increase over their squares simultaneously, so there is no other solution possible. QED.




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