Problem 25
Question
Solve each equation and check. \(2^{2 x+1}=16^{x}\)
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = \frac{1}{2}\).
1Step 1: Rewrite 16 as a Power of 2
The equation given is \(2^{2x+1}=16^{x}\). First, rewrite 16 as a power of 2. We know that \(16 = 2^4\). So, we can rewrite the right side as \((2^4)^x\).
2Step 2: Apply the Power of a Power Rule
Using the power of a power rule \((a^m)^n = a^{m \cdot n}\), we rewrite \((2^4)^x\) as \(2^{4x}\). The equation now becomes \(2^{2x+1} = 2^{4x}\).
3Step 3: Set Exponents Equal
Since the bases are the same (both sides have a base of 2), we can set the exponents equal to each other: \[2x+1 = 4x\]
4Step 4: Solve for x
Solve the equation \(2x + 1 = 4x\):Subtract \(2x\) from both sides:\[1 = 2x\]Divide both sides by 2:\[x = \frac{1}{2}\]
5Step 5: Verify the Solution
Substitute \(x = \frac{1}{2}\) back into the original equation to verify:Left side: \(2^{2(\frac{1}{2})+1} = 2^2 = 4\)Right side: \(16^{\frac{1}{2}} = (2^4)^{\frac{1}{2}} = 2^{4 \cdot \frac{1}{2}} = 2^2 = 4\)Both sides equal, so \(x = \frac{1}{2}\) is verified to be correct.
Key Concepts
Laws of ExponentsAlgebraic ManipulationPowers of Numbers
Laws of Exponents
The laws of exponents are a set of rules that make it easier to work with exponential expressions. They allow us to simplify and manipulate powers of numbers efficiently.
One key law is the **power of a power** rule, which states that
Another important law is the **same-base exponential equality**, which helps in solving equations like when both sides of the equation have the same base. We can safely say that the exponents themselves are equal:
One key law is the **power of a power** rule, which states that
- \((a^m)^n = a^{m \cdot n}\).
Another important law is the **same-base exponential equality**, which helps in solving equations like when both sides of the equation have the same base. We can safely say that the exponents themselves are equal:
- If \(a^m = a^n\), then \(m = n\).
Algebraic Manipulation
Algebraic manipulation involves rewriting and rearranging equations and expressions to uncover insights or find solutions. In the context of exponential equations, algebraic manipulation often involves:
In the given example, simple algebraic manipulations helped to equate both sides to a single base, uncovering the core equality through restating the problem in simpler terms.Steps forward often involve isolating variables or terms of interest, which provides clarity and facilitates efficient problem-solving.
- Changing the form of numbers, as seen in the exercise above where \(16\) was rewritten as \(2^4\).
- Applying laws of exponents to consolidate or simplify exponential terms.
In the given example, simple algebraic manipulations helped to equate both sides to a single base, uncovering the core equality through restating the problem in simpler terms.Steps forward often involve isolating variables or terms of interest, which provides clarity and facilitates efficient problem-solving.
Powers of Numbers
Understanding powers of numbers enables a deeper grasp of exponential relations. A power consists of a base and an exponent, organized in the form \(a^b\), where \(a\) is the base, and \(b\) is the exponent.
In this scenario:
This makes complex equations like \(2^{2x+1}=16^x\) much simpler to tackle, as it gives a base for comparison. The ability to adeptly manage powers of numbers is key to efficiently solving exponential equations.
In this scenario:
- Raising a number to a power is akin to multiplying that number by itself as many times as the exponent states.
- The base is the constant factor, while the exponent indicates the number of repetitions.
This makes complex equations like \(2^{2x+1}=16^x\) much simpler to tackle, as it gives a base for comparison. The ability to adeptly manage powers of numbers is key to efficiently solving exponential equations.
Other exercises in this chapter
Problem 25
In \(23-34,\) evaluate each function for the given value. Be sure to show your work. $$ f(x)=(2 x)^{-6} \div x^{3} ; f(-3) $$
View solution Problem 25
In \(3-37,\) express each power as a rational number in simplest form. $$ 7^{\frac{3}{4}} \times 7^{\frac{5}{4}} $$
View solution Problem 25
Show that if the area of one face of a cube is \(B,\) the volume of the cube is \(B^{\frac{3}{2}}\)
View solution Problem 25
Simplify each expression. In each exercise, all variables are positive. \(\frac{4(a b)^{2} c^{5}}{a b c}\)
View solution