Question

Use natural logarithms to solve each of the exponential equations. Hint: To solve \(3^{x}=11\), take \(\ln\) of both sides, obtaining \(x \ln 3=\ln 11 ;\) then \(x=(\ln 11) /(\ln 3) \approx 2.1827 .\) $$ 12^{1 /(\theta-1)}=4 $$

Short answer

\(\theta \approx 2.7925\)

Step-by-step solution

Take Natural Logarithm of Both Sides

We start by taking the natural logarithm (ln) of both sides of the equation to enable us to bring down the exponent in front of the log. The equation is \(12^{1/( heta-1)}=4\). Taking the natural log on both sides, we have: \(\ln(12^{1/( heta-1)}) = \ln(4)\).

Apply the Logarithm Power Rule

Using the power rule of logarithms, which states that \(\ln(a^b) = b \ln(a)\), we can bring the exponent \(1/( heta-1)\) in front of the natural logarithm. This gives us: \((1/( heta-1)) \ln(12) = \ln(4)\).

Solve for \(1/( heta-1)\)

We want to isolate \(1/( heta-1)\) on one side. We can do this by dividing both sides of the equation by \(\ln(12)\): \(1/( heta-1) = \ln(4)/\ln(12)\).

Solve for \(\theta\)

To find \(\theta\), recognize that \(1/( heta-1) = x\) where \(x = \ln(4)/\ln(12)\). Therefore, \((\theta-1) = 1/x = \ln(12)/\ln(4)\). Next, solve for \(\theta\) by adding 1 to both sides: \(\theta = \ln(12)/\ln(4) + 1\).

Compute the Numerical Solution

Finally, compute the values of \(\ln(12)\) and \(\ln(4)\), then find the quotient. Use a calculator or mathematical software to find: \(\theta \approx \frac{2.4849}{1.3863} + 1 \approx 2.7925\).

Key concepts

Natural Logarithms

A natural logarithm, denoted as \(\ln(x)\), is a logarithm with the base of \(e\), where \(e\) is an irrational number approximately equal to 2.71828. It is widely used in solving exponential equations. For example, when faced with an equation like \(12^{1/(\theta-1)}=4\), taking the natural logarithm on both sides allows us to manipulate the equation to help find the unknown variable \(\theta\).
This property arises because of the fundamental rule that:

• \(\ln(a^b) = b\ln(a)\)

This rule is pivotal as it enables us to "bring down" the exponent which facilitates solving for the variable.

Logarithm Properties

Logarithm properties are essential tools when working with exponentials. Among these properties, the logarithm power rule is extremely useful. It states that \(\ln(a^b) = b \ln(a)\). This allows us to transform equations in such a way that makes solving for unknowns much easier. For example, in the equation \(12^{1/(\theta-1)}=4\), applying the power rule gives us \((1/(\theta-1))\ln(12)=\ln(4)\).
Further properties of logarithms include:

• The product rule: \(\ln(ab) = \ln a + \ln b\)
• The quotient rule: \(\ln(a/b) = \ln a - \ln b\)
• The root rule: \(\ln(\sqrt{a}) = \frac{1}{2}\ln(a)\)

Understanding these properties aids in simplifying complex exponential equations into manageable forms.

Numerical Approximation

Numerical approximation is a method we use to find approximate values of expressions that are otherwise difficult to calculate by hand. This is particularly useful when dealing with logarithms. In our problem, calculating \(\ln(12)\) and \(\ln(4)\) precisely is challenging without a calculator, so we use approximation methods to find their values at approximately 2.4849 and 1.3863, respectively.
This method allows us to find approximate solutions for variables that meet real-world requirements, especially where precise values are not necessary. It’s also important for checking results derived from symbolic manipulations.

Exponentiation

Exponentiation is a mathematical operation involving two numbers: the base and the exponent. It is written as \(a^b\), where \(a\) is the base and \(b\) is the exponent. Solving exponential equations often involves isolating the base raised to a power. For example, in the equation \(12^{1/(\theta-1)}=4\), the goal is to solve for the variable in the exponent. This often requires manipulating the terms involved through logarithms.
The key when dealing with exponents is understanding the relationship between growth rates and exponent values, as well as how changing the exponent affects the outcome of the equation. Mastery of exponential functions and their manipulations through calculus and algebra helps immensely in solving complex problems.