math
posted by lisa .
solve each equation 1) log5log2x=1 2) 2log(x+1)=5

1)
log5  log(2x) = 1
log(5/2x) = 1
5/(2x) = 10^1
20x = 5
x = 1/4
2)
2log(x+1) = 5
log(x+1) = 5/2
10^(5/2) = x+1
316.227766 = x+1
x = 315.227766
OR
log(x+1)^2 = 5
(x+1)^2 = 10^5
x+1 = ±√(10^5) , but clearly x >1
x+1 = 316.227766..
x = 315.227766.. 
1)> log 5 + log 2 + log x = 1
>log (5 X 2) + log x = 1
>1 + log x = 1
>log x = 0
>x = 10^0
>x = 1
#2 has me twisted.