Question

If \(a=1+\log _{x} y z, b=1+\log _{y} 2 x\) and \(c=1+\log _{z} x y, \mathrm{t}_{d}\) \(a b+b c+c a\) is : (a) 1 (b) 0 (c) \(a b c\) (d) none of these

Short answer

Question: Given the following equations: 1. \(a = 1 + \log_x{yz}\) 2. \(b = 1 + \log_y{2x}\) 3. \(c = 1 + \log_z{xy}\) Find the value of the expression \(ab + bc + ca\). Answer: 3

Step-by-step solution

Understanding logarithm properties

To work with logarithms in these equations, we need to know few properties of logarithm: 1. \(\log_a{b} + \log_a{c} = \log_a{(b\cdot c)}\) 2. \(\log_a{b} - \log_a{c} = \log_a{(\frac{b}{c})}\) 3. \(a^{\log_a{b}} = b\)

Rewrite the equations using exponentials

Using the property 3 mentioned above, we can rewrite the given equations as exponentials: 1. \(x^a = yz\) 2. \(y^b = 2x\) 3. \(z^c = xy\)

Multiply the equations

Now we have a system of 3 equations with 3 variables. Let's multiply the exponentials, so we can simplify further: \(x^a \cdot y^b \cdot z^c = yz \cdot 2x \cdot xy = 2x^2y^2z^2\) This gives us: \((xyz)^{a+b+c} = 2x^2y^2z^2\)

Simplify

Apply logarithm on both sides of the equation: \(\log_{xyz}((xyz)^{a+b+c}) = \log_{xyz}(2x^2y^2z^2)\) Using the logarithm properties, we get: \((a+b+c)\log_{xyz}(xyz) = \log_{xyz}(2)+2\log_{xyz}(x)+2\log_{xyz}(y)+2\log_{xyz}(z)\) Since \(\log_{a}(a)=1\), we can rewrite the equation as: \(a+b+c = \log_{xyz}(2)+2\log_{xyz}(x)+2\log_{xyz}(y)+2\log_{xyz}(z)\)

Substitute and find the expression

We substitute the original expressions for \(a, b,\) and \(c\) back into the equation: \((1+\log_x{yz})+(1+\log_y{2x})+(1+\log_z{xy}) = \log_{xyz}(2)+2\log_{xyz}(x)+2\log_{xyz}(y)+2\log_{xyz}(z)\) Now, let's group the terms: \(3 + (\log_x{yz}+\log_y{2x}+\log_z{xy}) = \log_{xyz}(2)+2(\log_{xyz}(x)+\log_{xyz}(y)+\log_{xyz}(z))\) We want to find \(ab + bc + ca\), given by: \((1+\log_x{yz})(1+\log_y{2x})+(1+\log_y{2x})(1+\log_z{xy})+(1+\log_z{xy})(1+\log_x{yz})\) Notice that the desired expression will be equal to the left-hand side of the equation minus 3. So, we can rewrite the equation as: \((ab + bc + ca) = \log_{xyz}(2)+2(\log_{xyz}(x)+\log_{xyz}(y)+\log_{xyz}(z))-3\) To find the value of \(ab+bc+ca\), we need to simplify the right-hand side of the equation. We notice that both left-hand side and right-hand side are similar, with only a difference of 3. Therefore, the right-hand side of the equation minus the left side would be 3. \(ab+bc+ca = 3\) And the answer is (d) none of these.

Key concepts

Logarithmic Equations

Logarithmic equations involve variables within logarithms and are solved using properties of logarithms. Understanding these properties is crucial for both simplifying and solving logarithmic equations. To illustrate, let's consider the given exercise.

For example, the property \(\log_a{b} + \log_a{c} = \log_a{(b\cdot c)}\) allows us to combine logarithms with the same base, which is particularly useful when we need to simplify expressions or find the value of a variable. Similarly, \(a^{\log_a{b}} = b\) is a power property that demonstrates the inverse relationship between exponential and logarithmic functions.

In the exercise, properties like these are applied to manipulate and solve a complex problem involving multiple logarithms with different bases. The initial step toward solving logarithmic equations starts with rewriting them in an exponential form. This maneuver simplifies the expressions and turns a seemingly intricate series of logarithmic equations into more manageable exponential equations.

Quantitative Aptitude

In the context of quantitative aptitude, logarithms and exponentials frequently appear in problems meant to test numerical ability and problem-solving skills. They're not just mathematical concepts but tools that can help solve real-world problems involving growth rates, decay, and much more.

Quantitative aptitude questions often require a blend of concepts to arrive at the solution. This means grasping a wide array of mathematical tools—including arithmetic, algebra, and in particular, logarithms and exponentials—is essential. As seen in the given problem, quantitative aptitude challenges can combine these tools, requiring a step-by-step approach for solution.

After identifying the appropriate properties of logarithms and exponentials, we need to apply them correctly by substituting values and carrying out operations to simplify and find the right answer. This process demands careful attention to detail and a clear understanding of how different mathematical principles intersect.

Exponential Functions

Exponential functions can be defined by the equation \(f(x) = a^x\), where \(a\) is a positive constant. They are the inverse of logarithmic functions and depict growth or decay processes. In the context of the given problem, we manipulate exponential functions by using logarithms.

After rewriting the problem in exponential form, we see the equations transformed into \(x^a = yz, y^b = 2x, z^c = xy\). These functions are then multiplied, setting the stage for the application of logarithms to both sides of the resulting equation, which simplifies the equation to a single variable.

This reveals the interconnectedness of logarithms and exponential functions: understanding how to transition from a logarithmic equation to an exponential one, and vice versa, is a key skill in solving complex mathematical problems. Moreover, this task demonstrates how exponential growth patterns can be dissected into more understandable components through the use of logarithms, a technique that is widely used in various scientific and financial fields.