7^-(1/5) / 7^-(7/5) * 7^3/5 can be simplified as 7^a/b where a and b are positive integers.
find a*b-a
i'm not good when it come to negative exponent, plz help!
7^-(1/5) / 7^-(7/5) * 7^3/5
7^5/7(5/7) * 7 ^3/5
7^(8/5-3/5)=7*5/5
a=1, b=1 or whateer numbers as long as a=b.
This is too easy, if read it correctly.
a^(b/c)=a^(-c/b)
example
4^2=4^-1/2=1
7^5/7(5/7) * 7 ^3/5 <-- I don't understanding this part?
what is mean when the question ask find
a*b-a ?
I tries to figure out the answer you give but i just got more confuse.. where it 7^-(1/5)
To simplify the expression, we can use the property of negative exponents which states that a^(-n) is equal to 1/a^n.
Let's simplify each term in the expression:
1. 7^-(1/5) = 1 / 7^(1/5) (applying the negative exponent property)
2. 7^-(7/5) = 1 / 7^(7/5) (applying the negative exponent property)
3. 7^(3/5) (already in the correct form)
Now, let's substitute these values back into the expression:
(1 / 7^(1/5)) / (1 / 7^(7/5)) * 7^(3/5)
Using the division property of exponents, we can simplify: (a^m / a^n) = a^(m - n).
So, we have:
(1 / 7^(1/5)) * 7^(7/5) * 7^(3/5)
Now, let's apply the multiplication property of exponents: a^n * a^m = a^(n + m), and simplify:
= 1 * 7^(7/5 + 3/5)
= 7^(10/5)
= 7^2
Therefore, a = 2 and b = 1, and a * b - a = 2 * 1 - 2 = 0.