Problem 79 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \frac{\log x}{q-r}=\frac{\log y}{r-p}=\frac{\log z}{p-q}, \text { prove that } x^{q+r} \cdot y^{r+p} \cdot z^{p+q}=x^{p} \cdot y^{q} \cdot z^{r} . $$

Step-by-Step Solution

Answer

\(x^{q+r} \cdot y^{r+p} \cdot z^{p+q}= x^{p} \cdot y^{q} \cdot z^{r}\).

Step 1: Equate the given expression

Since the given equation states that \[\frac{\log x}{q-r} = \frac{\log y}{r-p} = \frac{\log z}{p-q},\] we can equate the individual pair of expressions: \[\frac{\log x}{q-r} = \frac{\log y}{r-p}\] and \[\frac{\log y}{r-p} = \frac{\log z}{p-q}.\]

Step 2: Cross-multiply and simplify

Cross-multiplying the first equation and simplifying, we get: \[\log x (r-p) = \log y (q-r).\] Now, cross-multiply the second equation and simplify: \[\log y (p-q) = \log z (r-p).\]

Step 3: Apply logarithm properties

Apply the logarithm property \(\log a^b = b \log a\) in both of the simplified equations: \[(r-p) \log x = (q-r) \log y\] and \[(p-q) \log y = (r-p) \log z.\]

Step 4: Divide both equations

Now, divide these two equations to get: \[\frac{(r-p) \log x}{(p-q) \log y} = \frac{(q-r) \log y}{(r-p) \log z}.\] Simplifying this equation, we get: \[\frac{r \log x - p \log x}{p \log y - q \log y} = \frac{q \log y - r \log y}{r \log z - p \log z}.\]

Step 5: Use logarithm properties again

Apply the logarithm property \(\log a - \log b = \log \frac{a}{b}\): \[\frac{\log \frac{x^r}{x^p}}{\log \frac{y^p}{y^q}} = \frac{\log \frac{y^q}{y^r}}{\log \frac{z^r}{z^p}}.\] Now, simplify using the property \(\log a = \log b\) if and only if \(a = b\): \[\frac{x^r}{x^p}\cdot \frac{y^p}{y^q} \cdot \frac{z^q}{z^r} = 1.\]

Step 6: Use the exponents rule

Apply the exponents rule \(\frac{a^m}{a^n} = a^{m-n}\) to simplify the expression: \[x^{r-p} \cdot y^{p-q} \cdot z^{q-r} = 1.\] Now, multiply both sides of the equation by \(x^p \cdot y^q \cdot z^r\): \[x^{q+r} \cdot y^{r+p} \cdot z^{p+q}= x^{p} \cdot y^{q} \cdot z^{r}.\] Hence, we have proved the required expression.

Key Concepts

Logarithm properties

The properties of logarithms are incredibly useful when simplifying complex equations and proving mathematical identities. One of the most important properties is the power rule for logarithms, which states that \( \log a^b = b \log a \). This means that you can bring an exponent in a logarithmic expression as a coefficient.

There are other essential properties as well:

• Product Property: \( \log(ab) = \log a + \log b \). This tells us that multiplying inside the logarithm translates to addition outside of it.
• Quotient Property: \( \log \left( \frac{a}{b} \right) = \log a - \log b \). Division inside the logarithm turns into subtraction outside it.
• Change of Base Formula: \( \log_b a = \frac{\log_c a}{\log_c b} \). This is particularly useful for converting logarithms from one base to another.

The exercise demonstrates the transformational power of these properties, helping to convert complex logarithmic expressions into simpler forms by manipulating their structure.

Exponent rules

Exponents are the backbone of many mathematical ideas, serving as a concise way to express repeated multiplication. One of the critical rules is the quotient rule, given by \( \frac{a^m}{a^n} = a^{m-n} \). This rule allows us to simplify expressions by subtracting exponents when dividing similar bases.

Here are other fundamental rules of exponents:

• Product Rule: \( a^m \times a^n = a^{m+n} \). Adding exponents is allowed when multiplying with the same base.
• Power Rule: \( (a^m)^n = a^{mn} \). Raising one exponent to another involves multiplying the exponents.
• Zero Exponent Rule: \( a^0 = 1 \), for any non-zero \( a \). This rule implies any non-zero number raised to the zero power is 1.

The solution of the exercise employs these rules to rearrange and simplify the given terms, proving a more complex expression by breaking it down using these fundamental exponent operations.

Cross-multiplication techniques

Cross-multiplication is a simple yet powerful technique used often in algebra to solve equations involving fractions. It involves multiplying the numerator of each fraction by the denominator of the other fraction, effectively eliminating the fractions.

Suppose you have an equation like \( \frac{a}{b} = \frac{c}{d} \). By cross-multiplying, you form the equation \( a \times d = b \times c \). This eliminates the fraction, often simplifying the process of solving for a variable.

In the solution to the original exercise, cross-multiplication helped equate portions of the expression to form non-fractional equations that could be further simplified. Once they were simplified, other algebraic techniques, like applying properties of logarithms and rules of exponents, were used to bring the proof to completion. Cross-multiplication provides a starting point where additional operations can subsequently lead to a solution.