Problem 42 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{7}\left(x^{3}+65\right)=0$$

Step-by-Step Solution

Answer

The solution is \( x = -4 \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_{7}(x^3 + 65) = 0 \). This means that we need to find the value of \( x \) such that when \( x^3 + 65 \) is raised to the base 7, it equals zero.

Step 2: Convert to Exponential Form

By definition, a logarithmic equation \( \log_b(y) = c \) can be rewritten in exponential form as \( y = b^c \). Here, the equation \( \log_7(x^3 + 65) = 0 \) can be rewritten as \( x^3 + 65 = 7^0 \).

Step 3: Simplify Exponential Equation

We know that any number raised to the power of zero is 1. Hence, \( 7^0 = 1 \). So, the equation simplifies to \( x^3 + 65 = 1 \).

Step 4: Solve for \( x \)

Subtract 65 from both sides of the equation to isolate \( x^3 \) on one side: \( x^3 = 1 - 65 \). Simplifying this gives \( x^3 = -64 \).

Step 5: Find the Cube Root

To solve for \( x \), take the cube root of both sides: \( x = \sqrt[3]{-64} \). The cube root of \(-64\) is \(-4\), as \((-4) \times (-4) \times (-4) = -64 \).

Step 6: Verify Solution with a Calculator

Using a calculator, verify that \( \,\log_{7}((-4)^3 + 65) = 0 \,\). Calculate \( (-4)^3 + 65 = -64 + 65 = 1 \) and \( \,\log_{7}(1) = 0 \,\), confirming the solution is correct.

Key Concepts

Exponential Form

Exponential form is a mathematical way to express repeated multiplication of a base number. It is the inverse operation of logarithms.
In a logarithmic equation, if we have \[\log_b(y) = c\]this can be rewritten in exponential form as \[y = b^c\].
Understanding this conversion is pivotal when solving logarithmic equations, like the one in the exercise we are discussing.

• In our exercise, we saw \[\log_7(x^3 + 65) = 0\] which was converted to \[x^3 + 65 = 7^0\].
• Since any number to the power of zero is always 1, the equation simplifies to \[x^3 + 65 = 1\].

Recognizing the relationship between exponential and logarithmic forms allows for easier manipulation of equations, offering a clear path to isolate variables and solve for unknowns.

Cube Roots

Cube roots provide a way to find the original base number that, when multiplied by itself twice, equals the given number.
This is represented as \[x = \sqrt[3]{a}\].
In our exercise, we encountered an equation \[x^3 + 65 = 1\], which simplified to \[x^3 = -64\].

• To solve for \(x\), take the cube root of both sides: \[x = \sqrt[3]{-64}\].
• The result, \((-4)\), confirms that multiplying \((-4)\) three times recreates the original form, \(-64\).

Understanding cube roots is essential in solving cubic equations, especially when simplifying equations to find the variable.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They help in solving equations where the variable is an exponent, making it easier to work with exponential growth or decay.
A basic logarithmic function is \[\log_b(a)\], determining the power to which base \(b\) must be raised to obtain \(a\).

• In our problem, the function \[\log_7(x^3 + 65) = 0\] directs us to find \(x\) by converting to exponential form, solving the underlying equation.
• Logarithms simplify complex multiplicative processes into additive ones, which are far easier to manipulate and solve.

Logarithmic functions are especially handy in real-life problems involving growth rates, acoustic decibels, and Richter scales, representing equations where changes occur exponentially.