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The quotient rule of logarithms is an important property that can simplify complex logarithmic expressions. It states that the logarithm of a quotient is equal to the difference of the logarithms:

• \( \log{\left(\frac{a}{b}\right)} = \log{a} - \log{b} \)

This rule helps transform subtraction of logs into a single logarithmic expression, making equations easier to solve. In the original exercise, this rule is used to combine \( \log{x} - \log{(x+1)} \) into the simpler expression \( \log{\left(\frac{x}{x+1}\right)} \).
By applying the quotient rule first, you consolidate the information and make the equation more manageable. This step is crucial before you can further transform the equation and solve for \(x\). Understanding and recognizing when to apply this rule is a helpful tool in tackling logarithmic equations.

Rewriting logarithmic equations in exponential form is a powerful technique to simplify and solve them. A logarithm equation like \( \log{\frac{x}{x+1}} = 2 \) reflects the relationship \( 10^b = a \), where \(b\) is the logarithm and the base is 10 (common logarithm). In this context, the equation becomes:

• \( \frac{x}{x+1} = 10^2 \)

This means \( \frac{x}{x+1} = 100 \), which is much simpler to work with. This form shows the direct ratio between \(x\) and \(x+1\), representing how logarithms establish a base-exponent relationship.
Switching to exponential form is essentially 'undoing' the logarithm. Recognizing this maneuver is essential for solving many logarithmic equations. Practicing the conversion from logarithmic to exponential form can significantly enhance problem-solving skills.

Solving equations, especially those involving logarithms, incorporates numerous steps that require methodical manipulation. After converting to exponential form, the exercise guides us to solve the equation

• \( \frac{x}{x+1} = 100 \)

by eliminating the fraction. Multiplying both sides by \( \x+1\) effectively clears the denominator, resulting in \( x = 100(x+1) \).
Distribute the 100 across \( (x+1) \) to yield \( x = 100x + 100 \). To isolate \( x \), move all \( x \)-terms to one side to obtain \( -99x = 100 \). The final piece involves solving for \( x \) by dividing both sides by -99, giving

• \( x = -\frac{100}{99} \).

These steps show the essential skills like distribution, combining like terms, and solving straightforward linear equations. Mastery of these skills will assist students in efficiently solving similar algebraic questions.