Solve the Natural Logarithm Equation: ln(4x) = Solution Steps

Q: Why can't I just use the property $ \ln(4x) = \ln 4 + \ln x $ ?

That property is correct for expanding logarithms, but this problem asks you to convert the entire expression $ \ln(4x) $ to base 7 format, not break it apart.

Q: How do I remember which log goes in the numerator vs denominator?

Think: "New base goes in denominator"! When converting to base 7, $ \log_7 e $ (the base 7 version of the original base e) goes on bottom.

Q: Is $ 4\ln x $ ever a correct answer for $ \ln(4x) $ ?

No! That would be $ \ln(x^4) $, not $ \ln(4x) $. The correct expansion is $ \ln 4 + \ln x $, but this problem wants base conversion instead.

Q: What if I need to convert to a different base like base 10?

Same formula! For base 10: $ \ln(4x) = \frac{\log_{10}(4x)}{\log_{10} e} $. Just replace the 7s with 10s in both the numerator and denominator.

Change of Base Formula with Natural Logarithms

$\ln4x=$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:03 We'll use the formula to convert from ln to log

00:06 We'll use this formula in our exercise

00:17 We'll use the formula for logarithmic division

00:22 We'll get the log of the numerator in base of the denominator

00:32 We'll use this formula in our exercise

00:47 We'll find the domain

01:02 We'll substitute 7

01:08 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$\ln4x=$

Step-by-step solution

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

Step 1: Identify the expression $\ln 4x$ .
Step 2: Apply the change-of-base formula to transform the natural logarithm to one using base 7.
Step 3: Use the formula $\ln a = \frac{\log_b a}{\log_b e}$ to substitute the values.

Now, let's work through each step:
Step 1: The expression given is $\ln 4x$ .
Step 2: We want a base 7 logarithm, so we apply the change-of-base formula:
Step 3: We have:

$\ln 4x = \frac{\log_7 4x}{\log_7 e}$

Therefore, the logarithmic expression $\ln 4x$ in base 7 is equivalent to $\frac{\log_7 4x}{\log_7 e}$ .

This matches the correct answer choice, which is choice 4.

Final Answer

$\frac{\log_74x}{\log_7e}$

Key Points to Remember

Essential concepts to master this topic

Formula: Use $\ln a = \frac{\log_b a}{\log_b e}$ to change base
Technique: Replace natural log with $\frac{\log_7 4x}{\log_7 e}$ for base 7
Check: Verify denominator is $\log_7 e$ not $\ln 7$ ✓

Common Mistakes

Avoid these frequent errors

Confusing natural log and base-7 log in denominator
Don't use $\ln 7$ in denominator = wrong base conversion! This mixes natural log notation with change-of-base formula incorrectly. Always use $\log_7 e$ in denominator when converting to base 7.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

What exactly is the change of base formula?

The change of base formula lets you convert any logarithm to a different base: $\log_a x = \frac{\log_b x}{\log_b a}$ . It's super useful when you need a specific base!

Why can't I just use the property $\ln(4x) = \ln 4 + \ln x$ ?

That property is correct for expanding logarithms, but this problem asks you to convert the entire expression $\ln(4x)$ to base 7 format, not break it apart.

How do I remember which log goes in the numerator vs denominator?

Think: "New base goes in denominator"! When converting to base 7, $\log_7 e$ (the base 7 version of the original base e) goes on bottom.

Is $4\ln x$ ever a correct answer for $\ln(4x)$ ?

No! That would be $\ln(x^4)$ , not $\ln(4x)$ . The correct expansion is $\ln 4 + \ln x$ , but this problem wants base conversion instead.

What if I need to convert to a different base like base 10?

Same formula! For base 10: $\ln(4x) = \frac{\log_{10}(4x)}{\log_{10} e}$ . Just replace the 7s with 10s in both the numerator and denominator.

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Solve the Natural Logarithm Equation: ln(4x) = Solution Steps

Change of Base Formula with Natural Logarithms

❤️ 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

What exactly is the change of base formula?

Why can't I just use the property $\ln(4x) = \ln 4 + \ln x$ ?

How do I remember which log goes in the numerator vs denominator?

Is $4\ln x$ ever a correct answer for $\ln(4x)$ ?

What if I need to convert to a different base like base 10?

🌟 Unlock Your Math Potential

More Questions

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Solve the Logarithmic Equation: log₄(x²+8x+1)/log₄8 = 2

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Solve the Natural Logarithm Equation: ln(4x) = Solution Steps

Change of Base Formula with Natural Logarithms

❤️ 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

What exactly is the change of base formula?

Why can't I just use the property ln⁡(4x)=ln⁡4+ln⁡x \ln(4x) = \ln 4 + \ln x ln(4x)=ln4+lnx?

How do I remember which log goes in the numerator vs denominator?

Is 4ln⁡x 4\ln x 4lnx ever a correct answer for ln⁡(4x) \ln(4x) ln(4x)?

What if I need to convert to a different base like base 10?

🌟 Unlock Your Math Potential

More Questions

Rules of Logarithms

Solve the Logarithm Ratio: Finding log₈(9a)/log₈(3a)

Solve the Reciprocal Expression: 1/ln8 Step-by-Step

Solve Logarithm Ratio: Finding log₄ₓ9/log₄ₓa Simplification

Solve the Logarithm Equation: x∙log_m(1/3^x)

Rules of Logarithms Combined

Solve 1/log₄(9): Simplifying Reciprocal Logarithm Expression

Solve the Inverse Logarithm: Finding (log₇x)⁻¹

Simplify the Expression: log₉(e²)/log₉(e) Using Log Properties

Solve the Logarithmic Equation: log₆8 Step by Step

Change-of-Base Formula for Logarithms

Solve the Logarithmic Equation: log₄(x²+8x+1)/log₄8 = 2

Solve for X: Log₈(4x) + Log₈(x+2) = 3Log₈(3) Equation

Solve log₇4: Calculate the Base 7 Logarithm Value

Solve the Logarithmic Fraction: log₈5 ÷ log₈9

Why can't I just use the property $\ln(4x) = \ln 4 + \ln x$ ?

Is $4\ln x$ ever a correct answer for $\ln(4x)$ ?