How do you solve the following math problem? 3^x + 4^x = 5^x?

Answer 1

It's clear that x = 2 is a solution, so our job is to show that there are no other solutions.

Divide both sides by 5^x :

(3^x + 4^x) / 5^x = 1

Separate the fraction:

(3^x)/(5^x) + (4^x)/(5^x) = 1

Distribute the exponent:

(3/5)^x + (4/5)^x = 1

I'll show that if x > 2, then x is not a solution to this equation.

Suppose x > 2. Since (3/5)^x and (4/5)^x are decreasing functions, then (3/5)^x < (3/5)^2 and (4/5)^x < (4/5)^2. Adding the two inequalities, we get

(3/5)^x + (4/5)^x < (3/5)^2 + (4/5)^2.

But we know the right side is just 1, so we know

(3/5)^x + (4/5)^x < 1.

So x cannot be a solution, because (3/5)^x + (4/5)^x is less than 1, and therefore is not 1.

Now I'll show that if x < 2, then x is not a solution.

Suppose x < 2. Since (3/5)^x and (4/5)^x are decreasing functions, then (3/5)^2 < (3/5)^x and (4/5)^2 < (4/5)^x. Adding these two inequalities gives us

(3/5)^2 + (4/5)^2 < (3/5)^x + (4/5)^x

But the left side is just 1, so we have

1 < (3/5)^x + (4/5)^x.

Since (3/5)^x + (4/5)^x is bigger than 1, then it's not 1, so x is not a solution.

So the only solution is x = 2.

Answer 2

First, as this (3,4,5) is a Pythagorean triple, we know that 3^2+4^2 = 5^2.

Fermat's Last Theorem states that there are no integer solutions to a^x+b^x = c^x, for x>2. a=3;b=4;c=5 is simply a specific case of this theorem. Thus, there can be no more solutions.

Answer 3

3^x+4^x-5^x= 0 so(3/5)^x+(4/5)^x-1=0
Take y=(3/5)^x+(/5)^x-1 Asx=>-infinity lim y= +infinity
as x=>+ infinity the limit is -1
y´= (3/5)^xln(3/5)+(4/5)^xln(4/5) As ln(3/5) and ln(4/5) are negative beeing 3/5 and 4/5 <1
y´is negative
so y is decreasing from +infinity to -1 and has only one root
which is x=2
Fermat´s last theorem speaks about integer solutions if x is a whole number >2 but doesn´t say anything if x is real number-.
If instead 3 4 and 5 you take 6,7,8 you´ll find also a solution
for x but not integer.

Answer 4

i'm going to respond to only the final question: 5(i) 4x -3> 3x -4 (-3x on the two facets) 4x -3x -3 > -4 x -3 > -4 (+3) x >-4 +3 x> -a million 5(ii) 5x +2 <4x +3 (-4x) 5x -4x +2 < 3 x +2 <3 (-2) x < 3-2 x < a million

Answer 5

x = 2
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Reason:
3-4-5 is a pythagorean triple.

Answer 6

x=2 pathagory theorem

its used for triangles that have one right angle

Answer 7

Using logs, I guess

Answer 8

x=2
i don't know how we can solve it but i just know it should be 2...

Answer 9

or study pythagorean theorem :)