challenging maths : log & α、β

There are few typo-errors, but I still understand.

Q1
x^2+2xlog5+log2.5=0
x = { -2log5 ± √ [(2log5)^2 - 4(1)(log2.5)] } / (2*1)
x = { -2log5 ± √ { 4[log(10/2)]^2 - 4log(10/4) } } / 2
x = { -2log5 ± √ [ 4(log10 - log2)^2 - 4(log10 - log4) ] } / 2
x = { -2log5 ± √ { 4(1 - log2)^2 - 4[1 - log(2^2)] } } / 2
x = { -2log5 ± √ { 4[1 - 2log2 + (log2)^2] - 4(1 - 2log2) } } / 2
x = { -2log5 ± √ [ 4 - 8log2 + 4(log2)^2 - 4 +8log2) ] } / 2
x = [ -2log5 ± √ (2log2)^2 ] / 2
x = ( -2log5 ± 2log2 ) / 2
x = - log5 ± log2

let α = - log5 + log2 and β = - log5 - log2
α = - log5 + log2
α = log2 - log5
α = log(2/5)

β = - log5 - log2
β = - ( log5 + log2)
β = - log(5*2)
β = - log10
β = - 1

10^α+10^β
= 10^[log(2/5)] + 10^( - 1)
= 2/5 + 1/10
= 1/2

Q2
貓貓(Commenter in area of comment) gives a link of solution of given equation,
but I am not familiar with W(product log function).
I choose linear approximation(using derivative) to do this general equation.
let A = 2^x
logA = log(2^x)
logA = x log2
d(logA)/dx = d(x log2)/dx
(1/A)(dA/dx) = log2
dA/dx = Alog2
d(2^x)/dx = (2^x)(log2)

y = 2^x+x
dy/dx = d(2^x+x)/dx
dy/dx = d(2^x)/dx + d(x)/dx
dy/dx = (2^x)(log2) + 1

[0 - y(x)] / (x ' - x) = dy/dx| x = x
- y(x) = (dy/dx| x = x)(x ' - x)
- y(x) = (dy/dx| x = x)(x ' ) - (dy/dx| x = x)(x)
(dy/dx| x = x)(x) - y(x) = (dy/dx| x = x)(x ' )
x - [ y(x) / (dy/dx| x = x) ] = x '
x ' = x - [ y(x) / (dy/dx| x = x) ]

how to use it?

x ' = x - [ y(x) / (dy/dx| x = x) ]
we choose x = 1 (any point is the same)
x ' = 1 - y(1) / (dy/dx| x = 1)
x ' = 1 - [ 2^(1)+(1) ] / { [2^(1)](log2) + 1 }
x ' = 1 - 3/(2log2+1)
x ' = - 0.87259 (5 sig fig)

then, calculate new x ' by replacing x = - 0.87259
x ' = - 0.87259 - [ 2^( - 0.87259)+( - 0.87259) ] / { [2^( - 0.87259)](log2) + 1 }
x ' = - 0.59226

repeating the steps
x ' = - 0.64119

- 0.64119 is the solution of 2^x+x = 0 (put x = - 0.64119 gives nearly 0)

2014-04-20 07:13:18 補充：
the graph helps you understand what actually the steps above happen

https://onedrive.live.com/redir?resid=A3980BFF0EA16013!202&authkey=!AB3tX9NCwGro_3g&v=3&ithint=photo%2c.png

2014-04-20 07:18:28 補充：
x ' = x - [ y(x) / (dy/dx| x = x) ]
x ' = x - { (2^x+x) / [(2^x)(log2) + 1] }
use calculator
press 1 and then EXE
solution = Ans - { (2^Ans + Ans) / [(2^Ans)(log2) + 1] }
repeating press EXE until the value does not change

2014-04-21 19:53:39 補充：
it is too trouble to input in long long equation.
it is very convenient.

http://www.wolframalpha.com/input/?i=2%5Ex%2Bx%3D0&t=crmtb01

2014-04-21 17:55:02 補充：
Excellent!

I like iterations!!!