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).