For what value of $a$ does $\left(\frac{1}{7}\right)^{3 a+3}=343^{s-1}$?
Answer (1)
Rewrite the equation using base 7: ( 7 − 1 ) 3 a + 3 = ( 7 3 ) a − 1 .
Simplify the exponents: 7 − 3 a − 3 = 7 3 a − 3 .
Equate the exponents: − 3 a − 3 = 3 a − 3 .
Solve for a : a = 0 .
Explanation
Rewrite with base 7 We are given the equation ( 7 1 ) 3 a + 3 = 34 3 a − 1 and asked to find the value of a that satisfies it. We can rewrite both sides of the equation using the base 7. Since 7 1 = 7 − 1 and 343 = 7 3 , we can rewrite the equation as ( 7 − 1 ) 3 a + 3 = ( 7 3 ) a − 1 .
Simplify exponents Now we simplify the exponents. Using the property ( x m ) n = x mn , we have 7 − 3 a − 3 = 7 3 a − 3 .
Equate exponents Since the bases are equal, we can equate the exponents: − 3 a − 3 = 3 a − 3 .
Solve for a Now we solve for a . Adding 3 a to both sides gives − 3 = 6 a − 3 . Adding 3 to both sides gives 0 = 6 a . Dividing both sides by 6 gives a = 0 .
Final Answer Therefore, the value of a that satisfies the equation is a = 0 .
Examples
Understanding exponential equations is crucial in various fields, such as finance and physics. For instance, in finance, compound interest can be modeled using exponential functions. If an investment grows according to the equation V = P ( 1 + r ) t , where V is the final value, P is the principal, r is the interest rate, and t is the time, solving for t might involve using logarithms, which are closely related to exponential equations. Similarly, in physics, radioactive decay is modeled using exponential functions, where the amount of a substance remaining after time t is given by N ( t ) = N 0 e − k t , where N 0 is the initial amount and k is the decay constant. Solving for t in this equation also involves logarithms and exponential equations. These examples demonstrate how understanding and solving exponential equations are essential in real-world applications.
