Solve the Logarithmic Equation: ln x + ln(x+1) - ln2 = 3

Logarithmic Equations with Exponential Solutions

Solve for X:

lnx+ln(x+1)ln2=3 \ln x+\ln(x+1)-\ln2=3

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:09 Let's find the value of X.
00:13 First, we'll apply the formula for combining terms, meaning we multiply them.
00:28 We'll use this multiplication formula in our exercise.
00:38 Next, we'll use a formula to subtract terms, dividing them as needed.
00:44 We'll apply this subtraction formula in our next steps.
01:04 Now, we'll use definitions to easily solve for the unknown.
01:19 To clear the fraction, we'll multiply through by 2.
01:29 Next, properly distribute each term inside the parentheses.
01:39 Let's carefully rearrange the equation to simplify.
02:04 We'll use the quadratic formula to find potential solutions.
02:19 It's time to calculate. Let's solve the expression.
02:44 Always remember, there might be two solutions: one with a plus, and one with a minus.
03:05 We'll determine the domain of definition next.
03:20 Here's the domain of definition. With it, we'll find the solution.
03:30 And that's the solution to our question. Great job!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve for X:

lnx+ln(x+1)ln2=3 \ln x+\ln(x+1)-\ln2=3

2

Step-by-step solution

The equation to solve is lnx+ln(x+1)ln2=3 \ln x + \ln(x+1) - \ln 2 = 3 .

Step 1: Combine the logarithms using the product and quotient rules:

ln(x(x+1))ln2=3becomesln(x(x+1)2)=3. \ln (x(x+1)) - \ln 2 = 3 \quad \text{becomes} \quad \ln \left(\frac{x(x+1)}{2}\right) = 3.

Step 2: Eliminate the logarithm by exponentiating both sides:

x(x+1)2=e3. \frac{x(x+1)}{2} = e^3.

Step 3: Solve for x x by clearing the fraction:

x(x+1)=2e3. x(x+1) = 2e^3.

Step 4: Expand and set up a quadratic equation:

x2+x2e3=0. x^2 + x - 2e^3 = 0.

Step 5: Use the quadratic formula x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=1 a = 1 , b=1 b = 1 , and c=2e3 c = -2e^3 :

x=1±124×1×(2e3)2×1. x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-2e^3)}}{2 \times 1}.

Step 6: Simplify under the square root:

x=1±1+8e32. x = \frac{-1 \pm \sqrt{1 + 8e^3}}{2}.

Step 7: Ensure x>0 x > 0 . Given 1+8e3 \sqrt{1 + 8e^3} will be positive, 1+1+8e32 \frac{-1 + \sqrt{1 + 8e^3}}{2} is the valid solution.

Therefore, the solution to the problem is 1+1+8e32 \frac{-1+\sqrt{1+8e^3}}{2} .

3

Final Answer

1+1+8e32 \frac{-1+\sqrt{1+8e^3}}{2}

Key Points to Remember

Essential concepts to master this topic
  • Rule: Combine logarithms using product and quotient properties first
  • Technique: Exponentiate both sides: ln(expression) = 3 becomes expression = e³
  • Check: Verify x > 0 and substitute back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting domain restrictions for logarithms
    Don't solve without checking x > 0 and x+1 > 0 = invalid solutions! Logarithms are only defined for positive arguments, so negative solutions must be rejected. Always verify that your answer makes all logarithmic expressions defined.

Practice Quiz

Test your knowledge with interactive questions

\( \log_{10}3+\log_{10}4= \)

FAQ

Everything you need to know about this question

Why can't I just move terms around like regular algebra?

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Logarithms have special properties that let you combine them first! Use ln a + ln b = ln(ab) and ln a - ln b = ln(a/b) to simplify before solving.

How do I get rid of the natural logarithm?

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Use the inverse operation! If ln(something) = 3, then something = e³. This works because exponential and logarithmic functions undo each other.

Why do I get a quadratic equation from a logarithmic equation?

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After eliminating the logarithm, you often get expressions like x(x+1) = 2e³. When you expand this, it becomes x² + x - 2e³ = 0, which is quadratic!

Do I need to check both solutions from the quadratic formula?

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Yes, always! Logarithms require positive arguments, so x > 0 and x+1 > 0. Only solutions that satisfy these conditions are valid.

What if my calculator doesn't have an e button?

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You can approximate e ≈ 2.718, so e³ ≈ 20.09. But for exact answers, leave it as e³ in your final expression.

How do I verify my answer is correct?

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Substitute your solution back into the original equation: ln x + ln(x+1) - ln 2 = 3. Calculate each term separately and check that they equal 3!

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