Problem 78 - Algebra and Trigonometry with Analytic Geometry

Question

(a) \(f(x)=\log \left(2 x^{2}+1\right)-10^{-x}, \quad x=1.95\) (b) \(g(x)=\frac{x-3.4}{\ln x+4}\), \(x=0.55\)

Step-by-Step Solution

Answer

(a) Approximately 2.12; (b) Approximately -0.84.

Step 1: Substitute into Function f(x)

To solve for \( f(1.95) \), we start by substituting \( x = 1.95 \) into the given function \( f(x) = \log(2x^2 + 1) - 10^{-x} \). This yields: \[ f(1.95) = \log(2(1.95)^2 + 1) - 10^{-1.95} \] First, calculate \( 2(1.95)^2 + 1 \): \[ 2(1.95)^2 = 2 \times 3.8025 = 7.605 \]\[ 7.605 + 1 = 8.605 \] Therefore, the expression becomes: \[ \log(8.605) - 10^{-1.95} \]

Step 2: Calculate Logarithm and Exponential Components of f(x)

Now, compute the natural logarithm and exponential components:- \( \log(8.605) \approx 2.13 \)- \( 10^{-1.95} \) can be computed using a calculator to get approximately \( 0.01122 \).Substitute these values back:\[ f(1.95) = 2.13 - 0.01122 \approx 2.11878 \]

Step 3: Substitute into Function g(x)

To solve for \( g(0.55) \), substitute \( x = 0.55 \) into \( g(x) = \frac{x - 3.4}{\ln(x) + 4} \). This yields: \[ g(0.55) = \frac{0.55 - 3.4}{\ln(0.55) + 4} \] Firstly, calculate the numerator:\[ 0.55 - 3.4 = -2.85 \]

Step 4: Calculate Logarithm Component of g(x)

Next, compute the natural logarithm and then the full denominator:- \( \ln(0.55) \approx -0.5978 \).Thus, the denominator becomes: \[ -0.5978 + 4 = 3.4022 \]So the function becomes:\[ g(0.55) = \frac{-2.85}{3.4022} \approx -0.8375 \]

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They answer the question: "To what power must the base be raised to get a certain number?" For example, in the logarithmic equation \( y = \log_b(x) \), \( b^y = x \) defines this relationship. Logarithms are commonly used in various domains like science, engineering, and finance.

When studying logarithmic functions, understanding the base is essential. The base \( b \) can be any positive number except 1, and different bases serve different purposes. Base 10 logarithms, or common logarithms, are widely used, while base \( e \), known as natural logarithms, is often applied in natural sciences.

Key Properties of Logarithmic Functions

• Product Property: \( \log_b(xy) = \log_b x + \log_b y \)
• Quotient Property: \( \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y \)
• Power Property: \( \log_b(x^a) = a \log_b x \)

Exponential Functions

Exponential functions describe situations where a quantity grows or decays at a rate proportional to its current value. Expressed in the form \( f(x) = a \cdot b^x \), they are characterized by a constant base \( b \) and an exponent \( x \).

These functions are profound in areas like population growth, radioactive decay, and compound interest calculations. The base \( b \) determines the nature of the exponential function:

• If \( b > 1 \), the function exhibits growth.
• If \( 0 < b < 1 \), the function displays decay.

Understanding Exponential Functions:
- Exponential Growth: Represented by functions where the amount increases over time, commonly found in environments with reproduction or multiplication.
- Exponential Decay: Often seen in processes diminishing over time like radioactive decay or depreciation in value.

Natural Logarithms

Natural logarithms have a unique and important base, the constant \( e \), which is approximately equal to 2.71828. These logarithms are denoted by \( \ln \). Used heavily in calculus, natural logarithms simplify a myriad of mathematical models.

A natural logarithm answers the question, "To what power must \( e \) be raised, to obtain a specific number?" This concept is vital in continuous growth or decay calculations, especially when involving compounding continuously.

Why Natural Logarithms Matter:
- They are used in defining the derivative and the integral of exponential functions.
- They have elegant properties when solving differential equations and modeling natural phenomena.

Common Properties:

• Inverse Property: \( e^{\ln x} = x \) and \( \ln(e^x) = x \)
• Derivative: The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \)