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solve, sqrt(11 - x) - sqrt(x + 6) = 3
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wait never mind I got it
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I have another problem too, a logarithm problem
solve 2^(x +1) = 7^(x + 2)
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Start: sqrt(11 - x) - sqrt(x + 6) = 3
Move sqrt(x + 6) to other side: sqrt(11 - x) = 3 + sqrt(x + 6)
Square both sides: [sqrt(11 - x)]^2 = (3 + sqrt(x + 6))^2 => 11 - x = (3 + sqrt(x + 6))^2
Expand RHS: 11 - x = 3^2 + 2*3*sqrt(x + 6) + sqrt(x + 6)^2
Simplify: 11 - x = 9 + 6 * sqrt(x + 6) + (x+6)
More: 11 - x = x + 15 + 6 * sqrt(x + 6)
More: 11 - x - x - 15 = 6 * sqrt(x + 6)
More: -4 - 2x = 6 * sqrt(x + 6)
Square both sides: (-4 - 2x)^2 = (6 * sqrt(x + 6))^2
Expand: (-4)^2 + -4*-2x + -2x*-4 + (-2x)^2 = 36 * (x+6)
Simplify: 16 + 2(8x) + 4x^2 = 36(x+6)
More: 16 +16x + 4x^2 = 36x + 216
More: 16-216 + 16x-36x + 4x^2 = 0
More: -200 -20x +4x^2 = 0
Divide by 4: x^2 - 5x - 50 = 0
Its a nice quadratic now!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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and so x = -5, 10 I got that to
now I just need help on the second problem
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2^(x +1) = 7^(x + 2)
Pulling out the known powers: 2^x * 2 = 7^x * 49
Then: 2/49 = 7^x / 2^x
Then: 2/49 = (7/2)^x
Using that special property of logarithms I cannot remember name of: log(2/49) = log(7/2) x
Rearranging: x = log(2/49)/log(7/2)
Calculator: x = -2.55
Or maybe neater this way:
2^(x +1) = 7^(x + 2)
Using that special property of logarithms I cannot remember name of: (x+1) log 2 = (x+2) log 7
expanding: x log 2 + log 2 = x log 7 + 2 log 7
pulling x's to one side: x log 2 - x log 7 = 2 log 7 - log 2
Then : x (log 2 - log 7) = 2 log 7 - log 2
Then: x = (2 log 7 - log 2) / (log 2 - log 7)
Calculator: x = 1.389 / -0.544 = -2.55
... test: 2^(x+1) = 7^(x+2) ==> 2^(-2.55+1) = 7^(-2.55+2) ==> 2^(-1.55) = 7^(-0.55) ==> 0.341 = 0.343 (close enough)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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thanks a bunch I hate logarithm stuff
The special property is i think just the property for logarithms with the same base:
where for positive mumbers b,x, and y where b != 1, log(b/x) = log(b/y) if and only if x = y.
I think...
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