If n and k are numbers greater than 1 and [tex]\( \sqrt[3]{n^4} \)[/tex] is equivalent to [tex]\( \sqrt[3]{k^3} \)[/tex], for what value of [tex]\( a \)[/tex] is [tex]\( n^{4a+1} = k \)[/tex]?
Answer (2)
We found that the value of a such that n 4 a + 1 = k is 12 1 . This was derived from equating the exponents after expressing k in terms of n . The relationship was established using the cube roots given in the problem.
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To solve the problem, let's logically break down the given expression and identify the relationship between the variables.
We are given that the cube root of n 4 is equivalent to the cube root of k 3 :
3 n 4 = 3 k 3
Since both are cube roots, we can equate their insides directly:
n 4 = k 3
This implies that:
k = ( n 4 ) 1/3 = n 4/3
So the value of k is n 4/3 .
Next, we need to find the value of a that satisfies the equation n 4 a + 1 = k . From our expression for k , we substitute:
n 4 a + 1 = n 4/3
Since the bases are the same ( n ), we can set the exponents equal to each other:
4 a + 1 = 3 4
To solve for a , first subtract 1 from both sides:
4 a = 3 4 − 1
4 a = 3 4 − 3 3
4 a = 3 1
Now, divide both sides by 4:
a = 3 1 × 4 1
a = 12 1
Therefore, the value of a is 12 1 .
