I need help in solving logarithm equations. Please Help!

1. Solve for x
log2 (x + 5) + log2 (x+2) = log2 (x + 6)

2. Solve for x
log (8x) - log (1 + sqrt x) = 2

Any work shown would be helpful.

1. log₂[(x+5)(x+2)] = log₂(x+6)
Raise both sides to the power of 2
(x+5)(x+2) = x+6
x² + x2 + 5x + 10 = x + 6
x² + 7x + 10 = x + 6
x² + 6x + 4 = 0

x = ( -b ± √[ b² - 4ac] ) / 2a
x = ( -6 ± √[ 6² - 4*1*4] ) / (2*1)
x = ( -6 ± √[ 36-16] ) / 2
x = ( -6 ± √20 ) / 2
x = ( -6 ± √(4*5) ) / 2
x = ( -6 ± √4√5 ) / 2
x = ( -6 ± 2√5 ) / 2
x = -3 ± √5
x = -3-√5, -3+√5
Now for the tricky thing. See the log(x+2) term? That means that x MUST be greater than -2, since you cannot get the log of a negative number. Since -3-√5 is less than -2 (and -3+√5 is more), the answer is x = -3+√5

2. log [8x/(1+√x)] = 2
Raise both sides to the base of 10
8x/(1+√x) = 10² = 100
Multiply both sides by (1+√x)
8x = 100(1+√x)
8x = 100 + 100√x
(8x - 100)/100 = √x
Square both sides:
x = [(8x-100)/100]²
x = (8x-100)²/100²
x = (64x² - 1600x + 10000) / 10000
10000x = 64x² - 1600x + 10000
64x² - 11600x + 10000 = 0
Divide throughout by 16
4x² - 725x + 625 = 0

x = ( -b ± √[ b² - 4ac] ) / 2a
x = ( -(-725) ± √[ (-725)² - 4*4*625 ] ) / (2*4)
x = ( 725 ± √[ 525625 - 10000 ] ) / 8
x = ( 725 ± √515625 ) / 8
x = ( 725 ± √(25 * 25 * 25 * 33) ) / 8
x = ( 725 ± √25√25√25√33 ) / 8
x = ( 725 ± (5*5*5)√33 ) / 8
x = ( 725 ± 125√33 ) / 8
x = ( 725 - 718.07... ) /8, ( 725 + 718.07...)/8
x = 0.8662..., 180.38379...

This time there's no problem with the range of x
However, when you substitute for each x in the original equation, the 180.38379 value works (log 100 = 2), while the 0.8662 one doesn't.

1. Solve for x
log2 (x + 5) + log2 (x+2) = log2 (x + 6)

log [base 2] [(x + 5)(x + 2)] = log [base 2] (x + 6)

so,

(x + 5)(x + 2) = (x + 6)

x^2 + 7x + 10 = x + 6

x^2 + 6x + 4 = 0

x = [- 6 ± √20]/2 = - 3 ± 2√5

check...

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