# How do you solve

41
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...

Bạn đang xem: How do you solve  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. 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. Highly active question. Earn 10 reputation in order lớn answer this question. The reputation requirement helps protect this question from spam và non-answer activity.

## Not the answer you're looking for? Browse other questions tagged calculus real-analysis or ask your own question.

Show that, for any constants \$a ∈ (0, 1)\$ and\$ b in hibs.vnbbR\$ the equation \$x = a sin x + b\$ has a chất lượng solution site design / hình ảnh © 2021 Staông chồng Exchange Inc; user contributions licensed under cc by-sa. rev2021.6.3.39424