solve for x; log(x-1) + log(x+1) =log(1 to the base2)
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Answered by
48
Given, log(x-1) + log(x+1) =log(1 to the base2)
We can write this as
log2(x-1) * (x+1) = 1
log2(x^2-1) = 1
x^2 - 1 = 1
x^2 = 2
x = root 2 and - root 2.
since log can be of negative value.
x = root 2.
Hope this helps!
We can write this as
log2(x-1) * (x+1) = 1
log2(x^2-1) = 1
x^2 - 1 = 1
x^2 = 2
x = root 2 and - root 2.
since log can be of negative value.
x = root 2.
Hope this helps!
Answered by
95
log 1 base 2 = 0
as 2^0=1.
Now, log(x-1) + log(x+1) =log(1 to the base2)
log(x-1) + log(x+1) = 0
log ( x ^ 2 -1) = 0
10^0 = x ^2 -1
1+1 = x ^2
x = √2
as 2^0=1.
Now, log(x-1) + log(x+1) =log(1 to the base2)
log(x-1) + log(x+1) = 0
log ( x ^ 2 -1) = 0
10^0 = x ^2 -1
1+1 = x ^2
x = √2
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