Expand the logarithm as much as possible. Rewrite the expression as a sum, difference, or product of logs.

 

ln(1/25^k)
 
 

Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example,  c*ln(h)
 .

 

ln(1/25^k)=
    

We begin by applying the logarithmic rules to expand the expression.

Given expression:

\[
\ln\left(\frac{1}{25^k}\right)
\]

### Step 1: Use the quotient rule of logarithms
The logarithmic rule for division is:

\[
\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)
\]

In this case:

\[
\ln\left(\frac{1}{25^k}\right) = \ln(1) - \ln(25^k)
\]

Since \(\ln(1) = 0\), we have:

\[
0 - \ln(25^k) = -\ln(25^k)
\]

### Step 2: Use the power rule of logarithms
The power rule states that:

\[
\ln(a^b) = b \cdot \ln(a)
\]

Applying this rule to \(\ln(25^k)\), we get:

\[
-\ln(25^k) = -k \cdot \ln(25)
\]

Thus, the fully expanded expression is:

\[
\ln\left(\frac{1}{25^k}\right) = -k \cdot \ln(25)
\]

---

Would you like more details on any of the steps, or do you have any further questions?

Here are 5 related questions to expand your understanding:

1. How do you simplify \(\ln(a \cdot b)\) using logarithmic properties?
2. What is the logarithm of any number raised to the power of zero?
3. How do you simplify expressions with logarithms of fractions?
4. Can you use the logarithmic expansion for trigonometric functions?
5. How would you expand \(\ln\left(\frac{a}{b^2}\right)\)?

**Tip:** Always remember that \(\ln(1) = 0\), which can help simplify expressions quickly!

Expand the logarithm as much as possible. Rewrite the expression as a sum, difference, or product of logs.

ln(1/25^k)

Learn how to expand the logarithmic expression ln(1/25^k) by applying the quotient and power rules of logarithms. This step-by-step solution covers core concepts like the quotient rule (ln(a/b)) and power rule (ln(a^b)), suitable for students in Grades 9-12.

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