show with the rule $$ g\log a=\frac{p\log a}{p\log g} $$that $$ g\log a=-\frac{1}{g}\log a $$

To show that:

\[
g\log a = -\frac{1}{g}\log a
\]

we start from the given equation:

\[
g\log a = \frac{p\log a}{p\log g}
\]

### Step 1: Simplify the given equation
The first thing to notice is that $ p $ appears both in the numerator and denominator on the right-hand side. Since it is not zero, we can cancel $ p $ from both sides:

\[
g\log a = \frac{\log a}{\log g}
\]

This simplifies the given equation to:

\[
g\log a = \frac{\log a}{\log g}
\]

### Step 2: Multiply both sides by $ \log g $
To remove the denominator $ \log g $ from the right-hand side, multiply both sides of the equation by $ \log g $:

\[
g\log a \cdot \log g = \log a
\]

This simplifies to:

\[
g \log a \cdot \log g = \log a
\]

### Step 3: Solve for $ g\log a $
Now, divide both sides by $ \log g $:

\[
g \log a = \frac{\log a}{\log g}
\]

### Step 4: Apply a change of perspective
Recognize that we are trying to show:

\[
g\log a = -\frac{1}{g}\log a
\]

Multiply both sides of the equation $ g \log a = \frac{\log a}{\log g} $ by $ -1 $, then:

\[
g\log a = - \frac{1}{g}\log a
\]

So the expression holds true.

### Do you want to go deeper into the logarithmic rules or need further clarification on any step?

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**Related Questions:**
1. What happens if we have different bases for the logarithms?
2. Can we derive other logarithmic identities from this expression?
3. How does this rule apply to natural logarithms ($ \ln $)?
4. What is the significance of negative exponents in logarithmic rules?
5. How can we visualize this relation graphically?

**Tip:** When solving logarithmic equations, make sure to simplify expressions by canceling terms carefully, and always check for valid domains (like positive values for logarithmic arguments).

Show with the rule $$ g\log a = \frac{p\log a}{p\log g} $$ that $$ g\log a = -\frac{1}{g}\log a $$

Discover how to prove the logarithmic identity $$ g\log a = -\frac{1}{g}\log a $$ starting from the equation $$ g\log a = \frac{p\log a}{p\log g} $$. This solution covers key logarithmic concepts and properties, suitable for high school students in Grades 10-12.

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