Math Problem Statement

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

Solution

To show that:

gloga=1glogag\log a = -\frac{1}{g}\log a

we start from the given equation:

gloga=plogaploggg\log a = \frac{p\log a}{p\log g}

Step 1: Simplify the given equation

The first thing to notice is that pp appears both in the numerator and denominator on the right-hand side. Since it is not zero, we can cancel pp from both sides:

gloga=logaloggg\log a = \frac{\log a}{\log g}

This simplifies the given equation to:

gloga=logaloggg\log a = \frac{\log a}{\log g}

Step 2: Multiply both sides by logg\log g

To remove the denominator logg\log g from the right-hand side, multiply both sides of the equation by logg\log g:

glogalogg=logag\log a \cdot \log g = \log a

This simplifies to:

glogalogg=logag \log a \cdot \log g = \log a

Step 3: Solve for glogag\log a

Now, divide both sides by logg\log g:

gloga=logaloggg \log a = \frac{\log a}{\log g}

Step 4: Apply a change of perspective

Recognize that we are trying to show:

gloga=1glogag\log a = -\frac{1}{g}\log a

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

gloga=1glogag\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?


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\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).

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Math Problem Analysis

Mathematical Concepts

Logarithmic Properties
Algebra

Formulas

$$ g\log a = \frac{p\log a}{p\log g} $$
$$ g\log a = -\frac{1}{g}\log a $$

Theorems

Logarithmic Identity
Properties of Logarithms

Suitable Grade Level

Grades 10-12