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Find all real solutions to the logarithmic equation ln (x) + ln (2) = 0?

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agleh | eNoter

Posted September 11, 2011 at 9:20 PM via web

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Find all real solutions to the logarithmic equation ln (x) + ln (2) = 0?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 11, 2011 at 9:23 PM (Answer #1)

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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

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giorgiana1976 | College Teacher | Valedictorian

Posted September 11, 2011 at 9:23 PM (Answer #2)

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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` .

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