the no. of solution of 2^x+3^x+4^x=5^x is
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hlo...the no. of solutions is 4
anisha49:
its wrong
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Say, f(x) = 2^x + 3^x + 4^x - 5^x = 0
For, x <=2, f(x) is positive.
For, x >=3, f(x) is negative.
So, there has to be at least one value of x that is between 2 and 3.
The value of x correct to 50 decimal places is
2.3732943684104698305305955807659755150751455928776
As function f(x) is monotonic, there cannot be more than one value of x between 2 and 3. Had it not been monotonic, there could have been 3, 5 or any odd number of values between 2 and 3.
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