×

# 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

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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