# math

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solve each equation 1) log5-log2x=1 2) 2log(x+1)=5

• math -

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..

• math -

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.

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