Problem 35 - Algebra A Combined Function

Question

Solve. $$ \left(\frac{1}{8}\right)^{x}=16^{1-x} $$

Step-by-Step Solution

Answer

The solution is \(x = 4\).

Step 1: Rewrite Bases in Terms of Powers of 2

First, notice that 8 can be rewritten as a power of 2: \[ \frac{1}{8} = (2^3)^{-1} = 2^{-3} \]Similarly, 16 can also be expressed as a power of 2:\[ 16 = 2^4 \] Therefore, the given equation can be rewritten as:\[ (2^{-3})^x = (2^4)^{1-x} \]

Step 2: Simplify the Exponents

Apply the property of exponents \( (a^m)^n = a^{m \cdot n} \) to both sides:\[ 2^{-3x} = 2^{4(1-x)} \] Now, equate the exponents because both sides have the same base:\[ -3x = 4(1-x) \]

Step 3: Solve the Equation for x

Expand the right side of the equation:\[ -3x = 4 - 4x \] Add \(4x\) to both sides to eliminate \(x\) on the right:\[ x = 4 \]

Step 4: Verify the Solution

Substitute \(x = 4\) back into the original equation to verify:\[ \left(\frac{1}{8}\right)^4 = 16^{1-4} \] Calculate both sides:- Left side: \( \left(\frac{1}{8}\right)^4 = \left(2^{-3}\right)^4 = 2^{-12} \) - Right side: \( 16^{-3} = (2^4)^{-3} = 2^{-12} \)Both sides are equal, confirming that \(x = 4\) is correct.

Key Concepts

Powers of 2

The concept of powers of 2 is critical in understanding and solving exponential equations, especially when the bases in the equation can be rewritten as powers of the same number. A number is expressed as a power of 2 when it can be represented as \(2^n\), where \(n\) is an integer. This is particularly useful in simplifying expressions and making the equation-solving process more manageable.

In the original exercise, we noticed that both 8 and 16 are related to powers of 2. Specifically:

• 8 is equivalent to \(2^3\), and so \(\frac{1}{8}\) becomes \(2^{-3}\).
• 16 is equivalent to \(2^4\).

Recognizing these relationships allows us to transform the bases in the given equation into the same base, namely 2, and significantly simplifies solving it. This transformation is crucial when you're working with exponential equations because it gives you a common ground (the base) to equate and solve using the properties of exponents.

Exponents

Exponents are a shorthand way to express repeated multiplication. Understanding how to manipulate exponents is essential for solving exponential equations. The general form \(b^a\) indicates that \(b\) (the base) is multiplied by itself \(a\) times. This compact way of writing large numbers helps in efficiently performing calculations and transformations.

In the exercise, using the property \((a^m)^n = a^{m \cdot n}\) helped us to simplify the expressions on both sides of the equation:

• \((2^{-3})^x = 2^{-3x}\)
• \((2^4)^{1-x} = 2^{4(1-x)}\)

This exponent property allows us to combine and equate expressions, making the math simpler. Understanding this and other exponent rules like \(a^m \times a^n = a^{m+n}\) and \(a^m / a^n = a^{m-n}\) broadens your ability to manipulate expressions where exponents are involved. Therefore, mastering the use of exponents and recognizing patterns in them is key to solving such mathematical problems.

Solving Equations

Solving equations, and particularly exponential equations, involves finding the value of the unknown variable that makes the equation true. In order to solve the given equation, we firstly required rewriting both sides with a common base. Once the equation was expressed in terms of powers of 2, it became possible to eliminate the base and work solely with the exponents.

The equation began as \(\left(\frac{1}{8}\right)^x = 16^{1-x}\), then transformed to \(2^{-3x} = 2^{4(1-x)}\). With both sides having the same base, it allowed us to directly equate the exponents: \(-3x = 4(1-x)\). This is a linear equation, which simplifies the process to solve for \(x\).

• We distributed on the right to get \(-3x = 4 - 4x\).
• By adding \(4x\) to both sides, we isolated \(x\), resulting in \(x = 4\).

Finally, substituting \(x = 4\) back into the original equation verified the solution, as both sides equaled \(2^{-12}\). Checking solutions by substitution is always crucial to ensure the accuracy and validity of the solution.