Problem 74 - College Algebra

Question

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth. $$ \ln (3)+\ln (4.4 x+6.8)=2 $$

Step-by-Step Solution

Answer

x ≈ -0.9857

Step 1: Use Logarithm Property

First, apply the property of logarithms: \( \ln(a) + \ln(b) = \ln(a \cdot b) \). Using this, we can combine the logarithms: \( \ln(3) + \ln(4.4x + 6.8) = \ln(3(4.4x + 6.8)) \). The equation now becomes: \( \ln(3(4.4x + 6.8)) = 2 \).

Step 2: Exponentiate Both Sides

To solve the equation, exponentiate both sides to remove the natural logarithms. This gives us: \( e^{\ln(3(4.4x + 6.8))} = e^2 \). Knowing that \( e^{\ln(y)} = y \), the equation simplifies to \( 3(4.4x + 6.8) = e^2 \).

Step 3: Simplify the Equation

Now, simplify the expression on the left: \( 3 \times 4.4x + 3 \times 6.8 = 13.2x + 20.4 \). So, the equation becomes: \( 13.2x + 20.4 = e^2 \).

Step 4: Solve for x

First, calculate \( e^2 \) using a calculator, which is approximately \( 7.3891 \). Now substitute this value back into the equation: \( 13.2x + 20.4 = 7.3891 \). Subtract 20.4 from both sides: \( 13.2x = 7.3891 - 20.4 = -13.0109 \). Divide by 13.2: \( x = \frac{-13.0109}{13.2} \).

Step 5: Round the Solution

Finally, use a calculator to perform the division: \( x = \frac{-13.0109}{13.2} \approx -0.9857 \). Round the answer to the nearest ten-thousandth to get \( x \approx -0.9857 \).

Key Concepts

Exponentiation

Exponentiation is a mathematical operation involving two numbers. The first is the base, and the second is the exponent. The concept of exponentiation means repeatedly multiplying the base by itself according to the exponent's value. For example, when we see an expression like \( 2^3 \), it means \( 2\times2\times2 \) which equals 8. In equation solving, exponentiation plays a crucial role in reversing logarithms. When you have logarithmic expressions in equations, raising both sides to the power of base \( e \) (the natural base for logarithms) can simplify your work. For example, in the step \( e^{\ln(y)} = y \), we used exponentiation to eliminate the natural logarithm \( \ln \) and work with \( y \) directly. With exponentiation, always remember:

• It's non-linear; results can grow rapidly or decline depending on whether the exponent is positive or negative.
• It aligns closely with real-world phenomena, like exponential growth in populations or radioactive decay.

Natural Logarithm

A natural logarithm is a special kind of logarithm where the base is the irrational number \( e \), approximately equal to 2.71828. We use the notation \( \ln(x) \) to represent natural logarithms. These logarithms are particularly handy in calculus and in solving growth and decay problems that arise naturally. Natural logarithms have some helpful characteristics:- They undo the effect of exponentiation with base \( e \). In formulas, \( e^{\ln(x)} = x \).- They are useful in solving equations where you need to "free" a variable from the exponent.- \( \ln(1) \) equals 0 as \( e^0 \) is 1. In solving the given exercise, using properties of natural logs simplifies complex expressions.If you're trying to solve an equation involving \( \ln \), think about exponentiating to \( e \) to simplify.

Properties of Logarithms

Logarithms are a fascinating way to "compress" multiplication, division, and exponentiation into addition or subtraction. The properties of logarithms are mathematical rules that help transform and simplify complex expressions.Some main properties include:

• Product Property: \( \ln(a) + \ln(b) = \ln(a \times b) \). This allows multiple logarithmic terms to be condensed into one, useful in reducing the complexity of expressions.
• Quotient Property: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This property helps in breaking down divisions within logarithms into a subtraction operation.
• Power Property: \( \ln(a^b) = b\ln(a) \). Here, an exponent outside the logarithm can be turned into a coefficient, aiding in calculus or derivations.

In our original exercise, the product property facilitated combining two logarithmic expressions into one, which was a vital step in moving forward to solve for \( x \). Knowing and applying these properties is essential in handling logarithmic equations efficiently.