Explain the error in the work shown. Find the correct answer.
$\begin{array}{l}
\frac{1}{64}=16^{2 a} \\
4^{-3}=\left(2^4\right)^{2 a} \\
4^{-3}=2^{8 a} \\
-3=8 a \\
-\frac{3}{8}=a
\end{array}$
Answer (1)
The original solution incorrectly equates exponents with different bases.
Rewrite both sides of the equation with the same base: 64 1 = 2 − 6 and 1 6 2 a = ( 2 4 ) 2 a = 2 8 a .
Equate the exponents: − 6 = 8 a .
Solve for a : a = 8 − 6 = − 4 3 . The correct answer is − 4 3 .
Explanation
Problem Analysis Let's analyze the given solution to identify the error and find the correct answer. The problem is to solve the equation 64 1 = 1 6 2 a for a . The provided solution attempts to rewrite both sides as powers of 2 or 4, but makes a mistake in equating exponents directly when the bases are different.
Original Equation The first step in the given solution is 64 1 = 1 6 2 a . This is the original equation we need to solve.
Rewriting with Powers of 4 and 2 The next step is 4 − 3 = ( 2 4 ) 2 a . Here, 64 1 is correctly rewritten as 4 − 3 since 4 3 = 64 , so 64 1 = 4 − 3 . Also, 16 is correctly rewritten as 2 4 .
Simplifying the Right Side The next step is 4 − 3 = 2 8 a . The right side, ( 2 4 ) 2 a , is correctly simplified to 2 8 a using the power of a power rule: ( x m ) n = x mn .
Identifying the Error The error occurs in the next step: − 3 = 8 a . The equation 4 − 3 = 2 8 a cannot be directly simplified to − 3 = 8 a because the bases are different (4 and 2). To equate the exponents, the bases must be the same. We need to rewrite 4 − 3 as a power of 2. Since 4 = 2 2 , we have 4 − 3 = ( 2 2 ) − 3 = 2 − 6 . So the correct equation should be 2 − 6 = 2 8 a .
Equating Exponents (Corrected) Now, with the same base, we can equate the exponents: − 6 = 8 a .
Solving for a Solving for a , we divide both sides by 8: a = 8 − 6 = 4 − 3 .
Final Answer and Conclusion Therefore, the correct answer is a = − 4 3 . The error in the original solution was equating the exponents when the bases were different.
Examples
Understanding exponential equations is crucial in many fields, such as calculating compound interest or modeling population growth. For example, if a population doubles every year and we want to know when it reaches a certain size, we use exponential equations. Similarly, in finance, understanding how investments grow over time involves solving exponential equations to determine the rate of return or the time it takes to reach a financial goal. These concepts are also fundamental in physics, such as in radioactive decay calculations.
