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log/ln probelm?
2lnx-(lnx)^2=0
Can someone explain how to do this problem
5 Answers
- llafferLv 74 years ago
You have:
2 ln(x) - [ln(x)]² = 0
We can factor out an ln(x) to create the product of two values that are equal to zero, so one of them must be equal to zero:
ln(x) [2 - [ln(x)] = 0
So now we have:
ln(x) = 0 and 2 - ln(x) = 0
x = 1 and ln(x) = 2
x = 1 and e²
- 4 years ago
Let ln(x) = t
2t - t^2 = 0
t^2 - 2t = 0
t * (t - 2) = 0
t = 0 , 2
ln(x) = 0 , 2
x = e^0 , e^2
x = 1 , e^2
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