(3^-3)^y(3^y)^-1 / 9^5 ,,,, 5^-x / (25^x)^2

Question

how to solve this two exponent step by step?

 

5^-x / (25^x)^2

 

(3^-3)^y(3^y)^-1 / 9^5

Answer 1

5^-x / (25^x)²

Lets simplify the denominator first. Raising a power to a power multiplies the exponents.

= 5^-x / 25^2x

We can write 25 as 5²

= 5^-x / (5²)^2x

= 5^-x / 5^4x

Now that the bases are the same, we can divide and subtract exponents

= 5^-x-4x

= 5^-5x

 

 

(3^-3)^y(3^y)^-1 / 9⁵

Again, we can multiply some exponents

= 3^-3y * 3^-y / 9⁵

Multiplying with the same base, you add exponents

= 3^-4y / 9⁵

Rewrite 9 as 3²

= 3^-4y / (3²)⁵

= 3^-4y / 3¹⁰

= 3^-4y-10