Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve for \(x\). (a) \(3 \ln x+5=(\ln x) \ln 2\) (b) \(2\left(7^{1+\log x}\right)=8\) (c) \(K e^{x}+K=L e^{x}-L\), where \(K\) and \(L\) are constants and \(0

Step-by-Step Solution

Answer

The solutions for each equation are \(x = \sqrt[3]{e^5/2}\), \(x = 10^{-3/7}\), \(x = \ln\left(\frac{L - K}{K - L}\right)\), \(x = \frac{\ln(R)}{n \ln(1+n) - \ln(Pn)}\) and \(x = \frac{\ln[3/(c/b)]}{\ln b - \ln c - \ln 3}\) respectively.

Step 1: Solve the first equation with logarithms

Let's start with the equation (a): \(3 \ln x+5=(\ln x) \ln 2\). First, distribute \(\ln 2\) to simplify: \(3 \ln x + 5 = \ln 2x\). Next, subtract \(\ln 2x\) from both sides to combine like terms: \(3 \ln x - \ln 2x + 5 = 0\). To solve for \(x\), use properties of logarithms to rewrite this as a single logarithm and exponentiate. The result is \(x = \sqrt[3]{e^5/2}\).

Step 2: Solve the second equation

Next, look at equation (b): \(2(7^{1+\log x})=8\). Divide both sides by 2: \(7^{1+\log x}=4\). Rewrite 4 as \(7^{4/7}\) to base match. This gives \(1 + \log x = 4/7\). Subtract 1 from both sides to get \(\log x = -3/7\). Finally, exponentiate both sides with base 10 to solve for \(x = 10^{-3/7}\).

Step 3: Solve the third equation

Now turn to equation (c): \(K e^{x}+K=L e^{x}-L\). This equation involves the exponential function \(e^{x}\). Start by getting terms with \(e^{x}\) on one side and the constant terms on the other: \(Ke^{x} - Le^{x} = L - K\). Factor out \(e^{x}\): \(e^{x}(K - L) = L - K\). Solve for \(x\) by dividing both sides by \(K - L\) and taking the natural logarithm: \(x = \ln\left(\frac{L - K}{K - L}\right)\).

Step 4: Solve the fourth equation

Next is equation (d): \(R(1+n)^{n x}=(P n)^{x}\). Simplify by dividing both sides by \(R\) and taking natural logs of both side, \(nx \ln(1 + n) = x \ln(Pn)\). Solve for \(x\) by dividing both sides by \([ n \ln(1 + n) - \ln(Pn) ]\) to obtain \(x = \frac{\ln(R)}{n \ln(1+n) - \ln(Pn)}\).

Step 5: Solve the fifth equation

Finally, equation (e): \(3 b^{x}=c^{x} 3^{2 x}\). Start by dividing both sides by \(3\) to obtain \(b^{x} = c^{x} 3^{x}\). Then take natural logs of both sides, \(\ln b^{x} = \ln c^{x} + x \ln 3\). Solve for \(x\) by dividing through by \([ \ln b - \ln c - \ln 3 ]\) to obtain \(x = \frac{\ln[3/(c/b)]}{\ln b - \ln c - \ln 3}\).

Key Concepts

Solving Exponential Equations

Solving exponential equations can seem daunting, but understanding the basics can greatly simplify the process. An exponential equation is one in which a variable appears in the exponent. To solve these, one common method involves getting the same base on both sides of the equation, which allows you to equate the exponents directly.

For example, consider the equation from the step-by-step solution, equation (b):
\(2(7^{1+\log x})=8\). The solution process involved rewriting 8 as \(7^{4/7}\) to match the base of 7, leading to \(1 + \log x = 4/7\). Once the bases are the same, it's simply a matter of solving for the variable within the exponent.

Properties of Logarithms

Logarithms are mathematical tools that allow us to work with exponential equations more easily, as they can turn multiplication into addition and division into subtraction. Key properties of logarithms include:

• The Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
• The Quotient Rule: \(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\)
• The Power Rule: \(\log_b(M^p) = p\cdot\log_b(M)\)

These properties are used in the provided step-by-step solutions to combine and simplify logarithmic terms, such as in equation (a) and equation (e). Understanding these properties can vastly improve a student's ability to manipulate and solve logarithmic equations.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. It appears in many areas of mathematics and science because of its unique properties related to growth processes.

For instance, in the step-by-step solution of equation (c), \(x\) is solved using the natural logarithm after isolating the exponential \(e^{x}\) term. Knowing when to use the natural logarithm is key, especially when dealing with continuous growth or decay problems or when the exponential function has base \(e\), as it simplifies the calculation.

Exponentiation

Exponentiation is the mathematical operation, written as \(b^n\), involving two numbers, the base \(b\) and the exponent \(n\). In essence, the exponent tells us how many times to multiply the base by itself. It's fundamental in solving exponential equations, because it allows us to move between the exponential form and a logarithm.

Understanding how to manipulate exponents is crucial, for example, in solving equation (d) of our exercise. Balancing the exponent terms and knowing the laws of exponents, such as \((a^b)^c = a^{b\cdot c}\) and \(a^b \cdot a^c = a^{b+c}\), can make finding the variable a much simpler task.