Solve Log₂x + Log₂(x/2) = 5: Finding the Value of x

Q: Why can't I just add the parts inside the logarithms?

Logarithms follow multiplication rules, not addition! When you add logarithms with the same base, you multiply their arguments: $ \log_b a + \log_b c = \log_b(a \times c) $. This is a fundamental logarithm property.

Q: How do I convert from logarithmic to exponential form?

If $ \log_b x = n $, then $ x = b^n $. In our problem, $ \log_2 x = 3 $ means $ x = 2^3 = 8 $. The base becomes the base of the exponent!

Q: Can x be negative in logarithmic equations?

No! The argument of a logarithm (the x inside $ \log_2 x $) must always be positive. That's why we don't consider negative solutions, even if they might satisfy the algebra.

Q: How do I check if x = 8 is really correct?

Substitute back: $ \log_2 8 + \log_2 \frac{8}{2} = \log_2 8 + \log_2 4 = 3 + 2 = 5 $ ✓. Since $ 2^3 = 8 $ and $ 2^2 = 4 $, our answer is correct!

Logarithmic Properties with Product Combination

$\log_2x+\log_2\frac{x}{2}=5$

?=x

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:09 Let's solve this problem step by step.

00:13 First, we'll find out where X can exist. This is called the domain.

00:26 To do this, we'll use a formula for adding logarithms. This gives us the logarithm of their product.

00:36 Let's apply this formula to our exercise now.

00:44 Now, we'll calculate the product as part of the process.

00:57 Next, we'll figure it out using the definition of logarithms.

01:02 To continue, we need to get X by itself, or isolate X.

01:11 When we extract a root, we find two solutions: one positive and one negative.

01:17 Let's see which solution fits with the domain we found earlier.

01:23 And there you have it! That's how we solve this question.

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$\log_2x+\log_2\frac{x}{2}=5$

?=x

Step-by-step solution

To solve this equation, we follow these steps:

Step 1: Use the property of logarithms $\log_b a + \log_b c = \log_b (a \cdot c)$ to combine terms on the left-hand side of the equation.
Step 2: Simplify the expression under the logarithm and solve for $x$ .

Let's proceed through these steps:

Step 1: Rewrite the equation using logarithmic properties:
$\log_2 x + \log_2 \frac{x}{2} = \log_2 x + \log_2 x - \log_2 2$

This simplifies to:
$2 \log_2 x - 1 = 5$

Step 2: Solve the equation:
Add 1 to both sides:

2 \log_2 x = 6

Divide both sides by 2:

\log_2 x = 3

Now, convert the logarithmic equation to its exponential form:

x = 2^3

Calculate $x$ :

x = 8

Therefore, the solution to the problem is $x = 8$ .

Final Answer

$8$

Key Points to Remember

Essential concepts to master this topic

Property: $\log_b a + \log_b c = \log_b(a \cdot c)$ combines logarithms
Technique: Simplify $\log_2 x + \log_2 \frac{x}{2} = \log_2\left(x \cdot \frac{x}{2}\right) = \log_2\left(\frac{x^2}{2}\right)$
Check: Substitute x = 8: $\log_2 8 + \log_2 4 = 3 + 2 = 5$ ✓

Common Mistakes

Avoid these frequent errors

Incorrectly applying logarithm properties
Don't add logarithms by adding their arguments: $\log_2 x + \log_2 \frac{x}{2} ≠ \log_2\left(x + \frac{x}{2}\right)$ = wrong result! This violates the logarithm addition rule and leads to impossible equations. Always multiply the arguments when adding logarithms: $\log_2 a + \log_2 b = \log_2(a \cdot b)$ .

Practice Quiz

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$\log_75-\log_72=$

FAQ

Everything you need to know about this question

Why can't I just add the parts inside the logarithms?

Logarithms follow multiplication rules, not addition! When you add logarithms with the same base, you multiply their arguments: $\log_b a + \log_b c = \log_b(a \times c)$ . This is a fundamental logarithm property.

How do I convert from logarithmic to exponential form?

If $\log_b x = n$ , then $x = b^n$ . In our problem, $\log_2 x = 3$ means $x = 2^3 = 8$ . The base becomes the base of the exponent!

Can x be negative in logarithmic equations?

No! The argument of a logarithm (the x inside $\log_2 x$ ) must always be positive. That's why we don't consider negative solutions, even if they might satisfy the algebra.

What if I get a different approach to solve this?

There are multiple valid approaches! You could use the product property like shown, or expand $\log_2 \frac{x}{2} = \log_2 x - \log_2 2$ . Both methods should give you x = 8 if done correctly.

How do I check if x = 8 is really correct?

Substitute back: $\log_2 8 + \log_2 \frac{8}{2} = \log_2 8 + \log_2 4 = 3 + 2 = 5$ ✓. Since $2^3 = 8$ and $2^2 = 4$ , our answer is correct!

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Solve Log₂x + Log₂(x/2) = 5: Finding the Value of x

Logarithmic Properties with Product Combination

❤️ Continue Your Math Journey!

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 can't I just add the parts inside the logarithms?

How do I convert from logarithmic to exponential form?

Can x be negative in logarithmic equations?

What if I get a different approach to solve this?

How do I check if x = 8 is really correct?

🌟 Unlock Your Math Potential

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