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Find all real solutions to the logarithmic equation ln (x) + ln (2) = 0?
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We have to find the real solutions of the logarithmic equation ln x + ln 2 = 0
Use the property log a + log b = log(a*b)
ln x + ln 2 = 0
=> ln(2*x) = 0
use the relation ln 1 = 0
=> ln (2*x) = ln 1
we can equate 2x and 1
=> x = 1/2
The solution of the equation is x = 1/2
Posted by justaguide on September 11, 2011 at 9:23 PM (Answer #1)
First, we'll impose the constraint of existence of ln(x):
x > 0
Since the logarithms from the equation have matching bases, we'll apply the product property of logarithms and we'll transform the sum of logarithms into a product.
ln (x) + ln (2) = ln (2x)
But ln (x) + ln (2) = 0 => ln (2x) = 0
We'll take antilogarithm and we'll have:
2x = `e^0`
2x = 1
x = `1/2`
Since the value of x is positive, then we'll accept it as solution of the given equation: x = `1/2` .
Posted by giorgiana1976 on September 11, 2011 at 9:23 PM (Answer #2)
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