A Famous Exponential

Divide both sides by $5^x$ $(\frac {3} {5}) ^ {x} +( \frac {4} {5} )^ {x} = 1$

We see that x = 2 is satisfying this equation, but since the left hand side has both fractions possessing larger denominator (Proper fractions) , we know that for all values of x greater than 2 the solution will never be 1. (Since it is a decreasing fraction)

Thus x=2 is the only solution.

If fractions don't seem clear enough, we can visualize it using decimals

$0.8^{x} + 0.6^{x} = 1$ Since only $0.64 + 0.36 = 1.00$ satisfy this equation, it is safe to say 2 is the only solution

We know that this is a Pythagorean Triple; thus, intuitively, it is easy to see that $x = 2$ . But, that doesn't constitute a ``proof" of this. Hence, we can exploit what we know about the above equation to confirm that there is no solution other than 2.

Let us insert the known solution, $x = 2$ :

$5^{2+x} = 4^{2+x} + 3^{2+x}$

$5^2 \cdot 5^x = (4^2 \cdot 4^x) + (3^2 \cdot 3^x)$

Dividing both sides by $5^x$ , we obtain:

$5^2 = 4^2 \cdot \left(\frac{4}{5}\right)^x + 3^2 \cdot \left(\frac{3}{5}\right)^x$

Now, if $x = 0$ , the equality holds. But, if we allow $x$ to increase, then the value of both summands on the right-hand-side will decrease below their squares, no longer holding. The same applies for decreasing $x$ ; both summands increase above their squares. Thus, $x$ must be zero in this case, leading to the conclusion that the solution is $x =2$ . QED

This question can be solved in two ways :-

1. The real method i.e Fermat's Last Theorem that any 3 numbers which are related as $a^x$ + $b^x$ = $c^x$ , x can be only 1 or 2 (i.e not greater than 2)

2. The side-kick method (the silly method btw) is the Pythagoras theorem. If you are a fan of Pythagoras theorem questions , you will immediately recognize the value of x , as it is common in many questions :D .

Fermat's Last Theorem states that no three positive integers $a$ , $b$ , and $c$ can satisfy the equation $a^{n} + b^{n} = c^{n}$ for any integer value of $n$ greater than $2$ .

Divide both sides by 5^x

Then the given equation can be written as

(0.6)^x + (0.8)^x = 1 ................... ( 1 )

x = 2 satisfies the given equation

i . e

(0.6)^2 + (0.8)^2 = 1

Let x be greater than 2

Then

(0.6)^x will be less than (0.6)^2 ...........(because 0.6 is a proper fraction)

also

(0.8)^x will be less than (0.8)^2............(because 0.8 is a proper fraction)

Then

(0.6)^x + (0.8)^x <(0.6)^2 + (0.8)^2

i . e

(0.6)^x + (0.8)^x < 1

which contradicts equation ( 1 )

So the given equation cannot be satisfied for x greater than 2 ...... ( I )

Now, let x be less than 2

Then

(0.6)^x will be greater than (0.6)^2 ...........(because 0.6 is a proper fraction)

Also

(0.8)^x will be greater than (0.8)^2............(because 0.8 is a proper fraction)

Then

(0.6)^x + (0.8)^x >(0.6)^2 + (0.8)^2

i . e

(0.6)^x + (0.8)^x > 1

which contradicts equation ( 1 )

So the given equation cannot be satisfied for x less than 2 ...... ( II )

From ( I ), ( II ) we get :

The solution set of the given equation is { 2 }

Solution is attached in the form of image

The wording of this question is confusing: the "sum of values of x"

Try substituting x with 2&3. Simple logic will tell us the answer

It's simple Pythagoras Maths, probably the most famous example of his work with triangles, the 3,4 and 5 Triangle. The answer is 2 as when you square 3 and 4 you get 9 and 16. When added together, the answer is 25, which is 5 squared (to the power of 2),

Because of Fermat's Last Theorem, $2$ is the only available answer.

By using fermat's last theorem X^n+y^n=z^n We can solve it So 3^2+4^2=5^2 So ans:2

Simple. Start assuming the values of x. Start with 2, 3²+4²=5² i.e- 9+16 which equals to 25 i.e 5². After assuming some more values you'll soon realise that 2 is the lone number that satisfies the given equation.

Because 5^x increases at an exponentially greater rate than 3^x or 4^x, we can plainly see that the only number that satisfies the equation is x=2. And for those of you saying it's 6, the question asks for the sum of all possible values of x, not sum of substitutions. Because the only possible value of x is 2, the sum is therefore the same.

Fermat's theorem. x must de 2.

This appears to be one of the Pythagorean Triads if you look at the diagram.

1. Pythagoras's Theorem
2. Fermat's Last Theorem.

This is Fermat's unsolved theorem, which has now been proven in a very complex mathematical paper. So while I know x=2 solves it, I also know no other values solve this.

How about -∞ as a solution for x??

This is just Pythagoras lol

The question asked for the sum of all the x's. Since x=2, sum should be 6. The answer to the question is 6 not 2

you asked for natural number... 0 is not a natural number :P

Take logs of everthing the simplify down so that x=(log(5))/(log(12))=1.943056387. So really the answer the author has put down is incorrect. But for arguments sake x=2

By substituting different values to satisfy the equation,we get answer

Fermat's last theorem can be used here It would have integral values only for x lesser than or equal to 2 . Checking all the values that are 0 , 1 , 2 , we see that x = 2 is the only solution . Therefore sum of all value(s) is 2

Is the Pitagorean theorem

3^2=9, 4^2=16, 9+16=25, Here x is replaced by 2, So, the value of X=2 .

3, 4 and 5 are pythagorean triplets,,,, H sq= P sq + B sq 5 tp 2= 3 tp2 + 4 tp2

It is a well known pythagorean triplet hence one answer is surely 2 while we know by Fermat's Theorem other natural values don't satisfy.......

Hope i'm right

Basically, we see the Pythagorean triple, 3-4-5, so 2 is indeed an answer. But, according to Fermat's Last Theorem, there are no solutions to x^n +y^n=z^n for n >2, so there are no more solutions

Simple, Pythagoras theorem. A^2+b^2=c^2

Given that $3^x+4^x=5^x$ $(3/5)^x+(4/5)^x=1$ $(0.6)^x+(0.8)^x=1$ $(1-0.4)^x+(1-0.2)^x=1$ $1-0.4 x+1-0.2 x=1$ $2-0.6x=1$ $1=0.6x$ $x=1.67$ This means that value of x should be near to 1.67 and rounding, it we get 2. Also as @Harsh Kumar said to confirm our answer the best way is Pythagorean theorem