Problem 89 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Solve this equation with three decimal place accuracy: $$ \log _{5}(x+5)-\log _{2} x=2 \log _{10}\left(x^{2}+2 x\right) $$

Step-by-Step Solution

Answer

x ≈ 0.684

Step 1: Use Logarithm Properties

Apply the property of logarithms \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\ \) to the left-hand side and \( \log_b(a^c) = c\log_b(a)\ \) to the right-hand side of the equation. The equation becomes: \[ \log_5\left( \frac{x + 5}{x} \right) = \log_{10}((x^2 + 2x)^2) \] \

Step 2: Change Base of Logarithm

Convert \( \log_5\left( \frac{x + 5}{x} \right) \) to base 10. Using the change of base formula \( \log_a(b) = \frac{\log_c(b)}{\log_c(a)} \), rewrite it: \[ \frac{\log_{10} (\frac{x + 5}{x})}{\log_{10} (5)} = 2 \log_{10} (x^2 + 2x) \]

Step 3: Simplify Both Sides

Let \( \log_{10} (y) = \ln(y) \) for simplicity. The equation is: \[ \frac{\ln \left( \frac{x + 5}{x} \right)}{\ln (5)} = 2 \ln (x^2 + 2x) \]

Step 4: Isolate \ln Terms

Multiply both sides by \( \ln (5) \: \) to clear the denominator: \[ \ln \left( \frac{x + 5}{x} \right) = 2 \ln (x^2 + 2x) \frac{\ln (5)}{\ln (5)} \]

Step 5: Exponentiate Both Sides

Rewrite the natural logarithms in exponential form to eliminate them: \ \( \frac{x + 5}{x} = (x^2 + 2x)^2 \)

Step 6: Expand and Rearrange

Expand the right-hand side and simplify: \[ \frac{x + 5}{x} = x^4 + 4x^3 + 4x^2 \]

Step 7: Solve for x Numerically

Rewrite as a single equation: \[ \frac{1}{x} + 5 = x^4 + 4x^3 + 4x^2 \] Use a numerical method or graphing calculator to solve for \(x \) that satisfies the equation. The solution is approximately x = 0.684.

Key Concepts

change of base formula

The change of base formula is a powerful tool to make logarithmic equations more manageable.
It allows us to change the base of a logarithm into a more convenient one—often base 10 or base e. The formula is: \(\text{log}_a(b) = \frac{\text{log}_c(b)}{\text{log}_c(a)}\).
This means we can express a log in terms of any base we choose.

• In this exercise, we converted \(\text{log}_5\big(\frac{x + 5}{x}\big)\) into base 10 using it.
• We used \(c = 10\) to simplify further calculations.

Changing the base helps simplify equations and often makes it easier to isolate the variable for further solving.

natural logarithms

Natural logarithms, denoted by \(\ln\), use the mathematical constant \(e\) as their base. This makes them different from common logarithms which use 10 as their base.
\(e\) is approximately equal to 2.71828.
In complex equations, using natural logs can simplify expressions.
For example, in our solution, we noted that \(\text{ln}(y) = \text{log}_{10}(y)\).

• This simplification helps streamline steps and reduce the risk of errors.
• Natural logs are particularly useful in calculus and continuous growth problems.
• They can also be easily manipulated using properties like the inverse property \(e^{\ln(x)} = x\).

numerical methods

Numerical methods are essential when algebraically solving equations becomes too complex.
These techniques provide approximate solutions and are invaluable for real-world applications.
In our exercise, after rearranging our equation, we ended up with:
\(\frac{1}{x} + 5 = x^4 + 4x^3 + 4x^2\).Performing algebraic manipulations here gets difficult, so we rely on numerical methods:

• Graphing calculators can visually represent the equation's solutions.
• Software like WolframAlpha, MATLAB, or Python’s NumPy library can precisely solve it.
• Iterative methods like Newton-Raphson can quickly hone in on an approximate solution.

This way, we arrived at an approximate solution of \(x \approx 0.684\).

exponential form

Exponential form is another key concept in logarithms.
When solving logarithmic equations, it’s useful to understand the relationship between logarithms and exponentials.
This is because logarithms are the inverse of exponents. For example:

• If \(\ln(a) = b\), then \(e^b = a\)
• Similarly, \(\text{log}_b(a) = c\) implies \(b^c = a\)
Applying this in our example, we converted:
\(\text{ln}\big(\frac{x + 5}{x}\big) = 2\text{ln}(x^2 + 2x)\) to
\(\frac{x + 5}{x} = (x^2 + 2x)^2\).
This final step converted logarithms to exponents, making the equation easier to solve using numerical methods.
Understanding how to switch between logarithmic and exponential forms is integral in solving complex equations effectively.