Problem 51 - Algebra 2

Question

Solve each equation. Round to four decimal places. $$ 4^{x}=\sqrt{5^{x+2}} $$

Step-by-Step Solution

Answer

x \approx 2.7666

Step 1: Rewrite the Equation

The original equation is \( 4^x = \sqrt{5^{x+2}} \). Recognize that \( \sqrt{5^{x+2}} \) can be rewritten as \( (5^{x+2})^{1/2} \). Therefore, the equation becomes \( 4^x = (5^{x+2})^{1/2} \).

Step 2: Simplify the Right Side

Apply the property of exponents to simplify the right side of the equation: \( (5^{x+2})^{1/2} = 5^{(x+2)/2} \). The equation now is \( 4^x = 5^{(x+2)/2} \).

Step 3: Equate Exponents Using Logarithms

Take the natural logarithm of both sides to equate the exponents: \( \ln(4^x) = \ln(5^{(x+2)/2}) \). This simplifies to \( x \ln(4) = \frac{x+2}{2} \ln(5) \).

Step 4: Solve for x

Reorganize to solve for \( x \):\[ x \ln(4) = \frac{x \ln(5)}{2} + \ln(25) \]Multiply through by 2 to clear the fraction: \[ 2x \ln(4) = x \ln(5) + 2 \ln(5) \]Rearrange to isolate x terms on one side: \[ 2x \ln(4) - x \ln(5) = 2 \ln(5) \]Factor out x:\[ x (2\ln(4) - \ln(5)) = 2\ln(5) \]Solve for x:\[ x = \frac{2\ln(5)}{2\ln(4) - \ln(5)} \]

Step 5: Calculate the Value of x

Use a calculator to compute the values: \( \ln(4) \approx 1.3863 \) and \( \ln(5) \approx 1.6094 \).Substitute these into the equation:\[ x = \frac{2(1.6094)}{2(1.3863) - 1.6094} \]Calculate:\[ x \approx \frac{3.2188}{2.7726 - 1.6094} \]\[ x \approx \frac{3.2188}{1.1632} \approx 2.7666 \].

Key Concepts

Logarithms

Logarithms are important tools in mathematics that help simplify expressions involving exponents. They allow us to "undo" exponentiation, just like square roots undo squaring a number. In an equation, logarithms can convert multiplicative and exponential relationships into additive ones, which are often easier to solve. Logarithms have a base, and the most common ones are base 10 (common logarithm) and base \(e\) (natural logarithm). For example, if we say \(\log_b(a) = c\), it means that \(b^c = a\).
Logarithms also have essential properties:

• Product Property: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
• Quotient Property: \(\log_b\left( \frac{M}{N} \right) = \log_b(M) - \log_b(N)\)
• Power Property: \(\log_b(M^p) = p\log_b(M)\)

Using these properties strategically can help solve equations involving logarithms and exponents.

Exponent Properties

When dealing with exponent equations, recognizing and applying the properties of exponents is crucial. Exponents tell us how many times to multiply a base by itself. Let's explore some key properties:

• Product of Powers: When multiplying two powers with the same base \(a\), add the exponents: \(a^m \cdot a^n = a^{m+n}\).
• Power of a Power: When raising a power to another power, multiply the exponents: \((a^m)^n = a^{mn}\).
• Power of a Product: Distribute the exponent over multiple bases in a product: \((ab)^n = a^n b^n\).
• Zero Exponent: Any non-zero base raised to the power of 0 is 1: \(a^0 = 1\).
• Negative Exponent: A negative exponent represents a reciprocal: \(a^{-n} = \frac{1}{a^n}\).

Understanding these can simplify complicated expressions and make solving exponential equations much more accessible.

Solving Equations

Solving equations with exponents can be challenging, but by breaking down the process into clear steps, it becomes manageable. Here’s how to tackle these problems effectively:
First, identify the equation type. Are the bases the same, or are logarithms involved? In cases involving exponential growth, equate the bases if they match, and use exponent rules if they don't.
Next, logarithms often come into play, as they help isolate the variable when applied correctly. Taking the logarithm on both sides can help to "bring down" exponents for easier manipulation. This is particularly useful when dealing with exponential equations, such as \(4^x = 5^{(x+2)/2}\).
By using the natural logarithm, you can simplify the equation into a linear form such as \(x \ln(4) - \frac{x+2}{2} \ln(5) = 0\), allowing the solution of \(x\) using algebraic methods. The next step involves isolating \(x\) by factoring and reorganizing, before calculating to find the numerical solution.

Natural Logarithm

Natural logarithms, denoted by \(\ln\), are logarithms with the base \(e\), where \(e\) is approximately 2.71828. This special logarithm is integral in calculus and exponential growth scenarios. It shares the same properties as regular logarithms, but it specifically deals with the natural exponent \(e\).
When solving equations, the natural logarithm converts exponential functions into linear ones. Applying \(\ln\) to an exponential equation like \(4^x = 5^{(x+2)/2}\), helps us equate exponents accurately: \(\ln(4^x) = \ln(5^{(x+2)/2})\). This converts the equation to \(x \ln(4) = \frac{x+2}{2} \ln(5)\).
Natural logarithms simplify expressions involving \(e\) and make solving for variables straightforward, without using calculator-specific logarithm bases. They are invaluable when handling growth and decay problems in mathematics and sciences, helping students see the relationship between exponential and logarithmic functions.