Problem 48 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.(0.1)^{2 \log 0.1-1.5 \log 0.1} \cdot(0.1)^{\left(\log \frac{8}{3}+2-\log 20\right)}\right) $$

Step-by-Step Solution

Answer

The simplified value of the given expression is \(\frac{4}{3}\).

Step 1: Rewrite the expression

Start by rewriting the expression to make it easier to work with. We can do this by multiplying the powers of the base (0.1) together: $$ \left(0.1^{\left(2\log 0.1-1.5\log 0.1\right)} \times 0.1^{\left(\log \frac{8}{3}+2-\log 20\right)}\right) $$

Step 2: Combine the exponents

Using the laws of exponents, we can combine the exponents by adding them. So the expression becomes: $$ 0.1^{\left(2\log 0.1-1.5\log 0.1+\log \frac{8}{3}+2-\log 20\right)} $$

Step 3: Simplify the exponents

Now, simplify the exponents using the properties of logarithms. Recall that \(\log a - \log b = \log\left(\frac{a}{b}\right)\) and \(\log a^b=b\log a\): $$ 0.1^{\left(\log(0.1^2) - \log(0.1^{1.5}) + \log \frac{8}{3}+2\log(0.1)-\log 20\right)} $$

Step 4: Apply logarithm properties

Use the properties of logarithms to further simplify the expression: $$ 0.1^{\left(\log \frac{0.1^2}{0.1^{1.5}}+\log \frac{8}{3}+2\log(0.1)-\log 20\right)} $$

Step 5: Simplify the fraction inside the logarithm

Simplify the fraction inside the logarithm by cancelling out the powers of 0.1: $$ 0.1^{\left(\log\frac{1}{0.1}+\log\frac{8}{3}+2\log(0.1) -\log 20\right)} \Rightarrow 0.1^{\left(\log 10 +\log \frac{8}{3}+2\log(0.1) -\log 20\right)} $$

Step 6: Combine logarithms

Combine the logarithms in the exponent: $$ 0.1^{\left(\log \left(10 \cdot \frac{8}{3} \cdot 0.1^2 \cdot \frac{1}{20}\right)\right)} $$

Step 7: Evaluate the expression inside the logarithm

Evaluate the expression inside the logarithm: $$ 0.1^{\left(\log \left(\frac{80}{60}\right)\right)} $$

Step 8: Simplify the fraction

Simplify the fraction: $$ 0.1^{\left(\log\frac{4}{3}\right)} $$

Step 9: Use the exponent property

Finally, use the exponent property where \(a^{\log_a(b)}=b\). Here, we have \(0.1^{\log(0.1)(\frac{4}{3})}\) which results in: $$ \frac{4}{3} $$ So, the simplified value of the given expression is \(\frac{4}{3}\).

Key Concepts

Exponentiation

Exponentiation is a mathematical operation that involves raising a number, which is called the base, to the power of an exponent. In simple terms, exponentiation refers to repeated multiplication of the base. For instance, if the base is 2 and the exponent is 3, it means you multiply 2 by itself three times: \[2^3 = 2 \times 2 \times 2 = 8\]In this context,

• "Base" is the number being multiplied.
• "Exponent" tells how many times to use the base in a multiplication.

Why is this important? In calculations involving logarithms, especially in our exercise, we often see bases like 0.1 being raised to various powers. It's crucial to understand how to manage these bases effectively by using the rules of exponentiation to combine and simplify expressions. Understanding exponentiation allows you to manipulate expressions more flexibly by combining or separating exponential terms, which is essential when tackling complex logarithm problems.

Laws of Logarithms

Logarithms are the inverse of exponentiation. They allow you to determine the power to which a number (the base) must be raised to get another number. The laws of logarithms are powerful tools that simplify this process. Here are three key laws:

• The Product Law: \( \log_b(xy) = \log_b(x) + \log_b(y) \) - It states that the log of a product is the sum of the logs.
• The Quotient Law: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \) - This indicates that the log of a quotient is the difference between the logs.
• The Power Law: \( \log_b(x^n) = n \times \log_b(x) \) - This law shows that the log of a number raised to an exponent is the exponent times the log of the base number.

These laws are instrumental when we're faced with expressions involving logarithms, like the one in our exercise. They simplify computations by converting multiplication, division, and exponentiation operations into addition, subtraction, and multiplication respectively with logs. Understanding these properties allows us to break down complex expressions into more manageable parts, facilitating easier calculations as seen throughout the step-by-step solution process.

Simplifying Expressions

Simplifying expressions means reducing them to a form that is simpler to work with, while maintaining equivalent value. In mathematical exercises, particularly those involving logarithms and exponentiation, simplifying expressions allows for clearer, more concise calculations.Whenever you simplify:

• Identify terms that can be combined using mathematical properties, like the power laws for logs and exponents.
• Look out for common factors or terms that can cancel out, streamlining the expression further.
• Apply simplification step by step to prevent errors and keep clarity, just as we broke down the logarithms and exponents in steps during the solution.

In our exercise, we made use of logarithmic and exponential properties to simplify the expression gradually. For example, by combining and simplifying logarithmic terms, we managed to reduce a complex logarithmic and exponential expression into something as straightforward as \( \frac{4}{3} \). Mastering the art of simplification is essential in making difficult equations and problems more approachable and in arriving at the correct solution efficiently.