Solve the Logarithmic Inequality: ln(x+5) + ln(x) ≤ ln(4) + ln(2x)

Logarithmic Inequalities with Domain Restrictions

x=? x=\text{?}

ln⁡(x+5)+ln⁡x≤ln⁡4+ln⁡2x \ln(x+5)+\ln x≤\ln4+\ln2x

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:13 Let's solve this problem together.
00:19 We will use the logarithm multiplication rule: find the log of the product.
00:30 Open parentheses, then multiply by each factor separately.
00:47 Now, let's set the logarithms equal to each other.
00:52 Time to arrange the equation neatly.
00:57 Factor out any common terms in the parentheses.
01:02 Let's determine the appropriate domain for this equation.
01:12 Examine the domain and find a solution that fits.
01:22 Great job! That's how we solve this question.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

x=? x=\text{?}

ln⁡(x+5)+ln⁡x≤ln⁡4+ln⁡2x \ln(x+5)+\ln x≤\ln4+\ln2x

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Use properties of logarithms to combine terms.

  • Step 2: Transform the logarithmic inequality into an algebraic form.

  • Step 3: Solve the resulting inequality.

  • Step 4: Check the domain restrictions and verify the solution.

Let's work through each step:

Step 1: Use the property ln⁡a+ln⁡b=ln⁡(ab) \ln a + \ln b = \ln(ab) :
ln⁡(x+5)+ln⁡x=ln⁡((x+5)x)=ln⁡(x2+5x) \ln(x+5) + \ln x = \ln((x+5)x) = \ln(x^2 + 5x)
ln⁡4+ln⁡2x=ln⁡(4⋅2x)=ln⁡(8x) \ln 4 + \ln 2x = \ln(4 \cdot 2x) = \ln(8x)

Step 2: Set up the inequality:
ln⁡(x2+5x)≤ln⁡(8x) \ln(x^2 + 5x) \le \ln(8x)

Step 3: Since the logarithmic functions are equal (i.e., both ordinals are decreasing or increasing simultaneously), we can drop logarithms (as long as the arguments are positive):
x2+5x≤8x x^2 + 5x \le 8x
Simplify the inequality to:
x2+5x−8x≤0 x^2 + 5x - 8x \le 0
x2−3x≤0 x^2 - 3x \le 0

Step 4: Factor the quadratic inequality:
x(x−3)≤0 x(x - 3) \le 0

Determine the critical points of the expression by setting each factor to zero:
x=0 and x=3 x = 0 \text{ and } x = 3

The critical points divide the number line into intervals: x<0 x < 0 , 0≤x<3 0 \le x < 3 , and x>3 x > 3 . Test these intervals:

  • For x<0 x < 0 , pick x=−1 x = -1 ; the expression (−1)(−1−3)=−4 (-1)(-1 - 3) = -4 , which is not less than or equal to zero.

  • For 0<x<3 0 < x < 3 , pick x=1 x = 1 ; the expression 1(1−3)=−2 1(1 - 3) = -2 , which satisfies the inequality.

  • For x>3 x > 3 , pick x=4 x = 4 ; the expression 4(4−3)=4 4(4 - 3) = 4 , which does not satisfy the inequality.

Finally, consider the endpoints:

  • At x=0 x = 0 , the inequality does not hold due to the logarithm constraints (undefined).

  • At x=3 x = 3 , substitute x x into the simplified inequality: 3(3−3)=0 3(3 - 3) = 0 , which satisfies the inequality.

Therefore, x x must satisfy the inequality 0<x≤3 0 < x \le 3 to maintain positive arguments for the logarithms and satisfy the inequality.

Thus, the solution to the problem is 0<x≤3 0 < x \le 3 , or choice 2.

3

Final Answer

0<X≤3 0 < X \le 3

Key Points to Remember

Essential concepts to master this topic
  • Logarithm Property: Use ln⁡a+ln⁡b=ln⁡(ab) \ln a + \ln b = \ln(ab) to combine terms
  • Technique: Drop logarithms when both sides positive: x2+5x≤8x x^2 + 5x \le 8x
  • Check: Verify domain restrictions: arguments must be positive and test endpoints ✓

Common Mistakes

Avoid these frequent errors
  • Ignoring domain restrictions when dropping logarithms
    Don't just drop the logarithms without checking that all arguments are positive = invalid solutions! Logarithms are only defined for positive arguments, so x>0 x > 0 , x+5>0 x + 5 > 0 , and 2x>0 2x > 0 must all be satisfied. Always verify domain restrictions before and after solving.

Practice Quiz

Test your knowledge with interactive questions

\( \log_75-\log_72= \)

FAQ

Everything you need to know about this question

Why can't I include x = 0 in my final answer?

+

Because logarithms are undefined at zero! The original equation has ln⁡x \ln x , and ln⁡(0) \ln(0) doesn't exist. Even though x(x−3)≤0 x(x-3) \le 0 includes x = 0, the domain restriction eliminates it.

How do I know when I can drop the logarithms?

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You can drop logarithms when both sides have the same base and all arguments are positive. Since ln⁡ \ln is an increasing function, ln⁡A≤ln⁡B \ln A \le \ln B means A≤B A \le B when both A and B are positive.

What if I forget to check the domain restrictions?

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You might include invalid solutions! Always check that arguments of logarithms are positive. For this problem, we need x>0 x > 0 , x+5>0 x + 5 > 0 , and 2x>0 2x > 0 , which gives us x>0 x > 0 .

Why is x = 3 included but x = 0 excluded?

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At x=3 x = 3 : all logarithm arguments are positive (ln⁡(8) \ln(8) , ln⁡(3) \ln(3) , ln⁡(6) \ln(6) ) and the inequality holds. At x=0 x = 0 : ln⁡(0) \ln(0) is undefined, so it's automatically excluded.

How do I test the intervals for the quadratic inequality?

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Pick a test point from each interval created by the zeros. For x(x−3)≤0 x(x-3) \le 0 with zeros at 0 and 3, test: negative values (like -1), values between 0 and 3 (like 1), and values greater than 3 (like 4). The inequality is satisfied when the product is negative or zero.

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