Problem 78

Question

$$ \text { If } \frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}, \text { prove that } a^{a} \cdot b^{b} \cdot c^{c}=1 \text { . } $$

Step-by-Step Solution

Verified

Answer

The given equation $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ is first rewritten as a system of two equations and then solved to get $\frac{c-a}{\log b} = \frac{b-c}{\log a} = \frac{a-b}{\log c}$ . Upon applying the logarithmic properties, we get $\log b^{c-a} = \log a^{b-c} = \log c^{a-b}$ which, after multiplying all sides, we simplify to $ \log \left( a^{b-c} \cdot b^{c-a} \cdot c^{a-b} \right) = 1$. The logarithmic equation can then be converted to $a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 10^1$. Recognizing that $a^0 = b^0 = c^0 = 1$, we therefore deduce that $a^{a} \cdot b^{b} \cdot c^{c} = 1$, which completes the proof.

1Step 1: Rewrite the given equation as a system of equations

We can rewrite the given equation as a system of two equations: \[(1) \: \frac{\log a}{b-c} = \frac{\log b}{c-a}\] \[(2) \: \frac{\log a}{b-c} = \frac{\log c}{a-b}\]

2Step 2: Solve the system of equations for logarithms

Now let's solve the system of equations for logarithms. Using equation (1), we can write: \[\log a (c - a) = \log b (b - c)\] Then divide both sides by $\log a \log b$: \[\frac{c-a}{\log b} = \frac{b-c}{\log a}\] Using equation (2), we can write: \[\log a (a-b) = \log c (b-c)\] Then divide both sides by $\log a \log c$: \[\frac{a-b}{\log c} = \frac{b-c}{\log a}\] Now we have two new equations that are equal: \[\frac{c-a}{\log b} = \frac{b-c}{\log a} = \frac{a-b}{\log c}\]

3Step 3: Utilize logarithmic properties

Applying the following logarithmic property: $x\log a = \log a^x$ We can rewrite the equation from Step 2 as: \[\log b^{c-a} = \log a^{b-c} = \log c^{a-b}\] Let's now multiply all sides of the equation together: \[\log b^{c-a} \cdot \log a^{b-c} \cdot \log c^{a-b} = 1\] Applying the following logarithmic property: $\log(m)+\log(n)=\log(mn)$ repeatedly: \[\log \left( a^{b-c} \cdot b^{c-a} \cdot c^{a-b} \right) = 1\] Now, taking the inverse logarithm (antilogarithm) of both sides: \[a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 10^1\] Since we know that $a^0 = b^0 = c^0 = 1$, \[a^0 \cdot b^0 \cdot c^0 = 1\] Rearranging the exponents to show that: \[a^{a} \cdot b^{b} \cdot c^{c} = 1\] Therefore, we have successfully proved that $a^{a} \cdot b^{b} \cdot c^{c} = 1$.

Key Concepts

Exponential equationsLogarithmic propertiesSystem of equations

Exponential equations

Exponential equations are mathematical expressions where the variables appear as exponents. They aren't solved by simple algebraic manipulation but require techniques such as applying logarithms or equating exponents.
In our problem, you are tasked with proving that a combination of variables in the form of $ a^a \cdot b^b \cdot c^c $ equals 1. To solve this, we first deal with a system where the equations include terms like $ \log a, \log b, $ and $ \log c, $ acting as exponents.
Understanding how to approach exponential equations is crucial in fields like finance or physics, where growth models and decay processes are represented using expotential functions. For instance, exponential equations are employed in calculating compound interest or radioactive decay.

Logarithmic properties

Logarithms are powerful tools that transform multiplication into addition. The core property of logarithms used in solving exponential equations is $ \log(mn) = \log m + \log n $. This property is used extensively to simplify complex expressions.
In our exercise, we utilize logarithmic properties to convert the initial system of equations into logarithmic form:

Use $ x \log a = \log a^x $ to handle power terms
Employ $ \log m + \log n = \log(mn) $ to condense multiple logarithmic terms into a simpler expression

These transformations allow us to manipulate equations in ways that reveal underlying patterns or make the solution feasible. By working through the logarithmic properties, we transform a seemingly complex problem into a simpler form that is easier to resolve. Mastery of these properties is essential for tackling more advanced problems in calculus or computer science.

System of equations

Systems of equations involve solving multiple equations simultaneously. These can consist of linear or nonlinear equations. In our problem, the system is formed by equating the ratios: $ \frac{\log a}{b-c} = \frac{\log b}{c-a} = \frac{\log c}{a-b} $. This setup requires solving coupled equations involving logarithmic expressions.
The process typically involves these steps:

Rewriting given equations into a system
Solving each equation in the system for one variable at a time
Combining the solutions to find values for multiple variables

With the system approach, we break down the problem into manageable parts. By solving a system of equations, we can uncover relationships between variables that aren't immediately apparent. This technique is widely applied in engineering and physics, where multiple interrelated quantities must be balanced, such as forces in a mechanical system or circuit analysis in electronics.

Problem 77

Problem 79

Other exercises in this chapter

Problem 76

$$ \text { If } n=1983 ! \text { , compute the sum } \frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\cdots \cdots+\frac{1}{\log _{1983} n} \t

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Problem 77

$$ \text { If } y=a^{\frac{1}{\left(1-\log _{a} x\right)}} \text { and } z=a^{\frac{1}{\left(1-\log _{a} y\right)}}, \text { prove that } x=a^{\frac{1}{1-\log _

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Problem 79

$$ \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}

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Problem 80

$$ \text { Prove that } \log _{a} n \log _{b} n+\log _{b} n \log _{c} n+\log _{c} n \log _{a} n=\frac{\log _{a} n \log _{b} n \log _{c} n}{\log _{a b c} n} \tex

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