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jasondelee · Feb 10, 2012, 10:04 AM

5^(x-2) 5^(x^2)=626 could u please solve the prob

corrigan · Feb 10, 2012, 12:44 PM

Can you verify that the problem is $5^{x-2}+5^{x^2}=626$ ? I'll help you, but as it stands, it's crazy hard.

ebaines · Feb 10, 2012, 12:48 PM

It's really not "crazy hard." For the equation

$ 5^{(x-2)} + 5^{x^2} = 626 $

you can tell by inspection that x must be greater than or equal to 2 - otherwise 5^(x-2) would be a fraction. So try a few values and see what you get.

corrigan · Feb 10, 2012, 12:57 PM

I got it. I was thinking of having e raised to both sides and using the cubic formula. I need to think more before I post.

jasondelee · Feb 10, 2012, 11:01 PM

you ,the problem is so.I've solved it through graph(but I ain't satisfied)and a way of pentanary (like binary).but the last way is applicable only when the answer is an integer.so all I need is an algebraic solution.could u please think about it.thanks a lot 4 replying.

ebaines · Feb 13, 2012, 09:56 AM

Jason:

I don't believe that there is a general solution to this type of problem, other than using numerical approximation techniques. For this specific problem it's pretty easy to guess at the answer. For a more formal approach one thing that you could try is setting up the right hand side as the sum of powers of 5. Note that 626 is quite close to 625, which is 5^4. So the equation can be rewritten as:

$ 5^{(x-2)} + 5^{x^2} = 626 = 5^4 + 5^0 $

So the answer is the solution that satisfies one of the following sets of equations - either:

$ x-2 = 4, \ and\\ x^2 = 0 $

or

$ x-2 = 0,\ and \\ x^2 = 4 $

There is no value for x that satisfies both equations of the first set, and the second set gives you x=2.