Solve the Logarithmic Equation: Log_4(3x^2+8x-10) = Log_4(-x^2-x+12.5)

Q: Why do I need to check if the expressions inside the logs are positive?

Logarithms are only defined for positive numbers! If you get x = 2 but plugging it in gives log(-3), that solution is invalid because you can't take the log of a negative number.

Q: What does it mean when log_4(something) = 0?

When $ \log_4(x) = 0 $, it means x = 4^0 = 1. Think of it this way: "4 to what power gives me the number inside the log?" If the answer is 0, then that number must be 1.

Q: How do I use the property log(a) - log(b) = log(a/b)?

When you have subtraction of logs with the same base, you can rewrite it as one log of a fraction. So $ \log_4(3x^2+8x-10) - \log_4(-x^2-x+12.5) = \log_4\left(\frac{3x^2+8x-10}{-x^2-x+12.5}\right) $

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

Once you set $ \frac{3x^2+8x-10}{-x^2-x+12.5} = 1 $, you cross-multiply to get linear terms and quadratic terms on both sides. Combining like terms naturally gives you a quadratic equation!

Q: What if one of my solutions doesn't work when I check it?

That's called an extraneous solution - it satisfies the quadratic but not the original logarithmic equation. Simply discard it and report only the valid solutions that make both log expressions positive.

Q: How do I check if 3x²+8x-10 > 0 for my solutions?

Substitute each solution: For x = 1.5: $ 3(1.5)^2 + 8(1.5) - 10 = 6.75 + 12 - 10 = 8.75 > 0 ✓ $ For x = -3.75: $ 3(-3.75)^2 + 8(-3.75) - 10 = 42.1875 - 30 - 10 = 2.1875 > 0 ✓ $

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:08 We want to find the domain

00:19 We'll use the roots formula to find the possible solutions

00:31 Let's calculate and solve

01:03 Remember that there are 2 possible solutions, positive and negative

01:34 These are the solutions, let's find the positive domain

01:46 This is the domain according to the first log

01:59 Now let's use the same method for the second log

02:04 We'll use the roots formula to find the possible solutions

02:14 Let's calculate and solve

02:34 Remember that there are 2 possible solutions, positive and negative

02:54 These are the solutions, let's find the positive domain

03:09 Now let's find the complete domain

03:24 This is the domain

03:44 We'll use the formula for log division, we'll get the quotient log

03:54 Let's use this formula in our exercise

04:19 Let's solve according to the log definition

04:24 Any number to the power of 0 always equals 1

04:29 Let's multiply by the denominator to eliminate the fraction

04:57 Let's arrange the equation

05:26 We'll use the roots formula to find the possible solutions

05:40 Let's calculate and solve

06:24 Let's note the 2 possible solutions

06:50 According to the domain, we'll find the solution

06:55 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

$\log_4(3x^2+8x-10)-\log_4(-x^2-x+12.5)=0$

?=x

2

Step-by-step solution

To solve this problem, we'll apply the following steps:

Step 1: Use logarithmic properties to rewrite $\log_4(3x^2+8x-10) - \log_4(-x^2-x+12.5) = 0$ as a single logarithm: $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$ .
Step 2: Recognize that if $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$ , then $\frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} = 4^0 = 1$ .
Step 3: Set up the equation: $3x^2 + 8x - 10 = -x^2 - x + 12.5$ .
Step 4: Rearrange the equation to: $3x^2 + 8x - 10 + x^2 + x - 12.5 = 0$ .
Step 5: Simplify to: $4x^2 + 9x - 22.5 = 0$ .
Step 6: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$ , $b = 9$ , $c = -22.5$ .
Step 7: Calculate: $b^2 - 4ac = 9^2 - 4 \times 4 \times (-22.5) = 81 + 360 = 441$ .
Step 8: Find $x$ : $x = \frac{-9 \pm \sqrt{441}}{8} = \frac{-9 \pm 21}{8}$ .
Step 9: Compute the roots: $x_1 = \frac{-9 + 21}{8} = \frac{12}{8} = 1.5$ and $x_2 = \frac{-9 - 21}{8} = \frac{-30}{8} = -3.75$ .
Step 10: Verify these solutions satisfy the domain of original logarithmic expressions by substituting back into $3x^2 + 8x - 10 > 0$ and $-x^2 - x + 12.5 > 0$ .

Therefore, the solutions to the problem are $x = -3.75, 1.5$ .

The correct choice from the provided options is: $-3.75, 1.5$ .

3

Final Answer

$-3.75,1.5$

Key Points to Remember

Essential concepts to master this topic

Rule: Use logarithm properties to combine logs: log(a) - log(b) = log(a/b)
Technique: If $\log_4(expression) = 0$ , then expression = 1
Check: Verify both solutions satisfy domain requirements for original logarithmic expressions ✓

Common Mistakes

Avoid these frequent errors

Forgetting to check domain restrictions
Don't solve the quadratic and assume both solutions are valid = wrong final answer! Logarithms require positive arguments, so some algebraic solutions might not work in the original equation. Always substitute both solutions back into the original expressions to verify they're positive.

Practice Quiz

Test your knowledge with interactive questions

$\log_{10}3+\log_{10}4=$

FAQ

Everything you need to know about this question

Why do I need to check if the expressions inside the logs are positive?

+

Logarithms are only defined for positive numbers! If you get x = 2 but plugging it in gives log(-3), that solution is invalid because you can't take the log of a negative number.

What does it mean when log_4(something) = 0?

+

When $\log_4(x) = 0$ , it means x = 4^0 = 1. Think of it this way: "4 to what power gives me the number inside the log?" If the answer is 0, then that number must be 1.

How do I use the property log(a) - log(b) = log(a/b)?

+

When you have subtraction of logs with the same base, you can rewrite it as one log of a fraction. So $\log_4(3x^2+8x-10) - \log_4(-x^2-x+12.5) = \log_4\left(\frac{3x^2+8x-10}{-x^2-x+12.5}\right)$

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

+

Once you set $\frac{3x^2+8x-10}{-x^2-x+12.5} = 1$ , you cross-multiply to get linear terms and quadratic terms on both sides. Combining like terms naturally gives you a quadratic equation!

What if one of my solutions doesn't work when I check it?

+

That's called an extraneous solution - it satisfies the quadratic but not the original logarithmic equation. Simply discard it and report only the valid solutions that make both log expressions positive.

How do I check if 3x²+8x-10 > 0 for my solutions?

+

Substitute each solution: For x = 1.5: $3(1.5)^2 + 8(1.5) - 10 = 6.75 + 12 - 10 = 8.75 > 0 ✓$ For x = -3.75: $3(-3.75)^2 + 8(-3.75) - 10 = 42.1875 - 30 - 10 = 2.1875 > 0 ✓$

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Solve the Logarithmic Equation: Log_4(3x^2+8x-10) = Log_4(-x^2-x+12.5)

Logarithmic Equations with Domain Verification

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Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why do I need to check if the expressions inside the logs are positive?

What does it mean when log_4(something) = 0?

How do I use the property log(a) - log(b) = log(a/b)?

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

What if one of my solutions doesn't work when I check it?

How do I check if 3x²+8x-10 > 0 for my solutions?

🌟 Unlock Your Math Potential

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