Problem 17 - College Algebra

Question

Solve each exponential equation in Exercises \(1-26\) Express the solution set in terms of natural logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$7^{0.3 x}=813$$

Step-by-Step Solution

Answer

The solution to the equation in terms of natural logarithms is \(x = \ln(813)/ (0.3 \cdot \ln(7))\). The decimal approximation, correct to two decimal places, will be calculated using a calculator.

Step 1: Convert the Exponential Equation to a Logarithmic Equation

Starting with the given equation \(7^{0.3x}=813\), we can rewrite this in logarithmic form using the definition of logarithms. Thus, the equation becomes \(0.3x = \ln(813)/\ln(7)\). This step leverages the relation between exponential and logarithmic forms which is \(b^y = x\) is equivalent to \(\log_b x = y\).

Step 2: Solve for the variable x

Now solve the logarithm equation for \(x\). It can be done by dividing both sides of the equation by 0.3. This gives us \(x = \ln(813)/ (0.3 \cdot \ln(7))\).

Step 3: Obtain the Decimal Equivalent

Now calculate the decimal equivalent of \(x\) by substituting the values of \(\ln(813)\) and \(\ln(7)\) using a calculator. This gives us the approximate value of \(x\).

Key Concepts

Natural Logarithms

Natural logarithms are fundamentally related to exponential functions, particularly with the base of Euler's number, denoted as 'e', where 'e' is approximately equal to 2.71828. The natural logarithm, commonly represented as \( \ln(x) \), is the power to which 'e' must be raised to obtain the value 'x'. It's an inverse function to the exponential function \( e^x \
\)For example, if \( e^3 = 20.085 \), then \( \ln(20.085) = 3 \). Natural logarithms are widely used in various scientific fields, including mathematics, physics, and engineering due to their unique properties and their natural occurrence in growth processes and compound interest calculations. In the context of solving exponential equations, natural logarithms are invaluable for finding the variable exponent when the base is 'e'.

Logarithmic Form

Understanding logarithmic form is key to manipulating and solving equations that feature exponents. Logarithms relate multiplicative and exponential changes to additive changes, which makes them particularly useful for solving equations where the variable is an exponent.
The logarithmic form is a way of expressing an exponential equation as a logarithm. For any positive numbers 'a', 'b', and 'x', if \( a^x = b \), then the logarithmic form is \( x = \log_a(b) \). Here, 'a' is called the base of the logarithm. This form highlights the exponent 'x' as the output of the logarithm of 'b' with base 'a'. In practical terms, logarithms allow us to bring exponents 'down to ground level', making them accessible for algebraic operations.

Exponential to Logarithmic Conversion

Converting an exponential equation to a logarithmic form is a crucial skill for solving these equations when the exponent is unknown. This process involves taking an equation in the form \( b^y = x \) and re-expressing it as \( y = \log_b(x) \). The base of the exponent becomes the base of the logarithm, while the outcome of the exponentiation is placed inside the logarithm.
For example, to convert the exponential equation \(7^{0.3 x} = 813\) to logarithmic form, we start by recognizing that 'x' is the unknown power to which 7 is raised to result in 813. The conversion would yield \(0.3x = \frac{\ln(813)}{\ln(7)}\), giving us an equation where 'x' can be isolated and solved using algebraic techniques.

Calculator Approximation

Once an algebraic expression for 'x' has been established using natural logarithms or common logarithms, obtaining a numerical approximation often requires the use of a calculator, as these values are not readily available through basic arithmetic. For a precise calculation, most scientific calculators have a dedicated \( \ln \) button, allowing you to find the natural logarithm of a number.
To approximate the solution obtained in logarithmic form, like \( x = \frac{\ln(813)}{(0.3 \cdot \ln(7))} \), you would input the natural logarithm of 813, divide it by the product of 0.3 and the natural logarithm of 7, and the calculator would output the value for 'x'. Rounding the result to the desired number of decimal places—usually two in textbook problems—provides an approachable, understandable answer for practical use.