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# Solution hunt

Find the number of solutions of 1) 2^x + 3^x + 4^x - 5^x = 0. 2) x^2 - 4 - [x] = 0 , Where [.] denotes the Greatest integer function. Have fun fellas and don't forget to provide an explanation!

Note by Sridhar Thiagarajan
4 years, 8 months ago

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write [x] = x - {x} now, x^2 - 4 = x - {x} or x^2-x-4 = -{x} now, -{x} belongs to (-1,0], hence find the intervals of values of x such that x^2 - x -4 also belongs to(-1,0] you will find that the values are aproxx [-1.56,-1.3) or (2.3,2.56] hence the values are such that [x] = -2 or 2, if [x] = -2, x= -( 2)^1/2, and if [x] = 2, x = (6)^1/2 Hence, 2 SOLUTIONS

- 4 years, 8 months ago

1) $$2^x + 3^x + 4^x = 5^x$$

Suppose x is large. Then dividing both sides by $$5^x$$ makes $$(\frac{2}{5})^5+(\frac{3}{5})^5+(\frac{4}{5})^5=1$$

This makes the LHS a strictly decreasing function, RHS stays constant as 1, hence there is only 1 solution.

Obviously 2<x<3. With a calculator (which I think there is another method) it is easy to get a close approximation of the answer.

- 4 years, 8 months ago

the second line raises the fraction to 5, but it should be x.

- 4 years, 8 months ago

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