Can we prove that an equation has only one solution?

[Math Tutor]

Can we prove that x = 2 is the only one solution of the equation

3^x + 4^x = 5^x

?

[dduclam]

Can we prove that x = 2 is the only one solution of the equation

3^x + 4^x = 5^x

?

Yes,we can. Put [tex]f(x)= (\frac3{5})^x+(\frac4{5})^x-1[/tex]
we have [tex]f'(x) >0[/tex]
=> [tex]f(x)=0[/tex] have only root!

[Guest]

It should be easier to prove that the equation has one root only.
Why f'(x) > 0 ?

[Guest]

Since the derivative is always positive, the function is always increasing. The graph crosses the x-axis at x= 2. Since the graph is always "going up", it can never get down to 0 again.

I don't know why you say "It should be easier to prove that the equation has one root only." That was pretty darn easy!

[Guest]

Perhaps the OP hasn't taken Calculus and does not know what "f'(x)" means.

[Guest]

My problem is with the simple statement that f'(x)>0 with no proof.
Since [tex]f(x)= \left(\frac{3}{5}\right)^x+ \left(\frac{4}{5}\right)^x- 1[/tex],
[tex]f'(x)= \left(\frac{3}{5}\right)^xln(3/5)+ \left(\frac{4}{5}\right)^xln(4/5)[/tex].
Both ln(3/5) and ln(4/5) are negative (3/5 and 4/5 are less than 1), it appears to me that f' is always negative!

[HallsofIvy]

But if f'(x) is always negative, it still follows that the equation has only one root!