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x=2 is one solution, and there's another somewhere in (-3,-2).
Why did the vector cross the road?
It wanted to be normal.
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mathsy,
what is the method to find such solutions?
TI-89 says the two solutions are:
x = 2
x = -ln(250)/ln(10)
Last edited by LuisRodg (2008-11-11 09:01:46)
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Given that
2^(x+2) 5^(6-x) = 10^(x^2)
= (2 x 5)^(x^2)
=2^(x^2) 5^(x^2)
Compare the LHS and RHS
x + 2 = x^2 and 6 - x = x^2
x^2 - x - 2 = 0 x^2 + x - 6 = 0
x = 2, -1 and x = 2, -3
This is our required answer.
Last edited by Prakash Panneer (2008-11-11 09:34:45)
Letter, number, arts and science
of living kinds, both are the eyes.
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On the integers it's easy to see that the RHS of the equation is a power of 10, which means you need to have the same number of 2's and 5's on the LHS. It just so happens that the value of x at which this occurs makes the entire equation true.
Non-integer solutions are more tricky. You need to do some fancy footwork with logarithms:
Now you can use the quadratic formula to solve for x.
The + half turns out to be x = 2. Whether or not the - half matches your TI-89's answer is beyond me, as I'm not sure how to simplify the expression further.
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What my answer is right only.
for example...
a^3 = a^x
We can take x = 3, since the bases are equal.
Similarly we can solve the given problem.
Hence the answer would be...
2^(x+2) 5^(6-x) = 10^(x^2)
= (2 x 5)^(x^2)
=2^(x^2) 5^(x^2)
Compare LHS and RHS
x + 2 = x^2 and 6 - x = x^2
x^2 - x - 2 = 0 x^2 + x - 6 = 0
x = 2, -1 and x = 2, -3
This is our required answer.
Last edited by Prakash Panneer (2008-11-11 10:37:11)
Letter, number, arts and science
of living kinds, both are the eyes.
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The + half turns out to be x = 2. Whether or not the - half matches your TI-89's answer is beyond me, as I'm not sure how to simplify the expression further.
As long as you're convinced that the + half is 2, you can use that fact to express the - half more simply.
Not sure if that helps, but it's better than before.
Why did the vector cross the road?
It wanted to be normal.
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Yep, it's easy to pull LuisRodg's second answer from that:
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