Math Lesson 13.4.5 - Equations Involving Natural Logarithm

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Welcome to our Math lesson on Equations Involving Natural Logarithm, this is the fifth lesson of our suite of math lessons covering the topic of Natural Logarithm Function and Its Graph, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Equations Involving Natural Logarithm

We can solve the equations involving natural logarithms in the same way as the rest of logarithmic functions. Thus, first we isolate the term(s) containing the variable, then we use any exponential transformation to isolate the variable only and eventually calculate its value. Let's consider a pair of examples to clarify this point.

Example 2

Solve the following equations.

1. e^x/3 - 1 = 1/e³
2. ln (10x) = ln 3 + ln (4x - 2)
3. ln (x² + 3) - ln x = ln 4

Solution 2

1. At the first sight, this seems an exponential equation. However, we cannot solve it without using the natural logarithm. We have
   e^x/3-1 = 1/e³
   e^x/3 - 1/e³ = 1
   e^x/3 - e^-3 = 1
   Multiplying both sides by e³ yields
   e³ ∙ (e^x/3 - e^-3 ) = e³ ∙ 1
   e³ ∙ e^x/3 - e³ ∙ e^-3 = e³
   e^{3 + x/3} - e^{3 - 3} = e³
   e^{9/3 + x/3} - e⁰ = e³
   e^{9 + x/3} - 1 = e³
   e^{9 + x/3} = e³ + 1
   Taking the 'ln' of both sides yields
   ln (e^{9 + x/3}) = ln (e³ + 1)
   (9 + x/3) ∙ ln e = ln (e³ + 1)
   (9 + x/3) ∙ 1 = ln (e³ + 1)
   9 + x/3 = ln (e³ + 1)
   9 + x = 3 ln (e³ + 1)
   x = 3 ln (e³ + 1) - 9
   x = 3 ln (e³ + 1) - 9
   x = 3 ln (20 + 1) - 9
   x = 3 ln 21 - 9
   x = 3 ∙ 3.045 - 9
   x = 9.135 - 9
   x = 0.135
2. ln (10x) = ln 3 + ln (4x-2)
   ln (10x) = ln 3 ∙ (4x-2)
   ln (10x) = ln (12x - 6)
   Now that we have a single 'ln' term in both sides, we can focus on the arguments only. Thus,
   10x = 12x - 6
   6 = 12x - 10x
   6 = 2x
   x = 3
3. ln (x² + 3) - ln x = ln 4
   ln (x² + 3)/x = ln 4
   Now that we have a single 'ln' term in both sides, we can focus on the arguments only. Thus,
   (x² + 3)/x = 4
   x² + 3 = 4x
   x² - 4x + 3 = 0
   This is a quadratic equation where the coefficients and the constant are a = 1, b = -4 and c = 3. The discriminant Δ is positive, as
   ∆ = b² - 4ac
   = (-4)² - 4 ∙ 1 ∙ 3
   = 16 - 12
   = 4
   Therefore, this equation has two distinct roots:
   x₁ = -b - √∆/2a
   = -(-4) - √4/2 ∙ 1
   = 4 - 2/2
   = 2/2
   = 1
   and
   x₁ = -b + √∆/2a
   = -(-4) + √4/2 ∙ 1
   = 4 + 2/2
   = 6/2
   = 3
   Sometimes, none of the roots in a natural logarithmic equation are accepted, as the argument becomes negative for that value. For example, solving the following equation
   ln (2x) + ln x = ln 18
   yields
   ln (2x ∙ x) = ln 18
   ln 2x² = ln 18
   2x² = 18
   x² = 9
   x = √9
   x = -3 and x = 3
   We reject the first root, as ln (2x) = ln [2 · (-3)] = ln (-6) and ln x = ln (-3). Both these logarithms are undefined, as they are negative. Therefore, we accept only the value x = 3 as a solution for this equation.

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7. Continuing learning logarithms - read our next math tutorial: Definition and Properties of Logarithms

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