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Differential Calculus

Blazing goIITian

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9 Aug 2009 22:23:38 IST

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Find the no. of solutions of 2^x + 3^x + 4^x - 5^x = 0

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Find the no. of solutions of 2^x + 3^x + 4^x - 5^x = 0

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Comments (9)

Akshay

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9 Aug 2009 22:32:10 IST

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no. of soln is 1...

Tanveer Sarma

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9 Aug 2009 22:37:39 IST

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I know the answer plz give me the solution.

Akshay

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9 Aug 2009 22:53:25 IST

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2^x+3^x+4^x=5^x

now assume 2^x+3^x+4^x=f(x)

5^x=g(x)

draw the graph of both functions and see the no. of points where they both intersect....

if this does not help let me know plz...

Tanveer Sarma

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9 Aug 2009 22:58:34 IST

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How to draw the graph of 2^x + 3^x + 4 ^x????

Akshay

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9 Aug 2009 23:05:42 IST

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f(x)>0 always....derivative is greater than 0 means monotonic inc.

now check values of f(x)at x=0,1,2,3 which is not difficult...u will see graps of both funcn intersect bw 2 and 3 and then for x>3 graph of f(x) is always below graph of g(x)...

Tanveer Sarma

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9 Aug 2009 23:09:12 IST

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Thanx.......

dc16

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9 Aug 2009 23:12:20 IST

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u can also compare both the graphs using superimposition and finding the value at 0 .

adi

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9 Aug 2009 23:20:04 IST

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ya divesh's answer is da best

vijay kharya

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16 Jun 2010 19:35:37 IST

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also we can solve it by this method, take 5^x to the RHS and then divide by 5^x, we get (2/5)^x+(3/5)^x+(4/5)^x=1, now this graph is made by 3 decresing functions which will intersect the straight line y=1 only once, hence only one solution

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