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#1 2005-05-15 11:57:04

im really bored
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Registered: 2005-05-12
Posts: 76

Another problem :(

solve,    sqrt(11 - x)   -   sqrt(x + 6)  = 3

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#2 2005-05-15 12:01:11

im really bored
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Registered: 2005-05-12
Posts: 76

Re: Another problem :(

wait never mind I got it

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#3 2005-05-15 12:02:49

im really bored
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Registered: 2005-05-12
Posts: 76

Re: Another problem :(

I have another problem too, a logarithm problem

solve  2^(x +1) = 7^(x + 2)

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#4 2005-05-15 12:15:29

MathsIsFun
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Registered: 2005-01-21
Posts: 7,535

Re: Another problem :(

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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#5 2005-05-15 12:35:04

im really bored
Member
Registered: 2005-05-12
Posts: 76

Re: Another problem :(

and so x = -5, 10  I got that to

now I just need help on the second problem

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#6 2005-05-15 12:44:22

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Another problem :(

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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#7 2005-05-15 15:30:51

im really bored
Member
Registered: 2005-05-12
Posts: 76

Re: Another problem :(

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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