Written By AI
Might be long.
Step-by-step explanation: 1. Evaluate (25-^(3/2))^(1/3):
To evaluate this expression, we'll start by simplifying the innermost exponent, which is 3/2. The square root of 25 is 5, so we can rewrite the expression as (25^0.5)^(1/3).
Next, we'll apply the exponent 1/3 to 25^0.5. This means we need to find the cube root of 25^0.5. The cube root of 25 is 2, so the final result is 2.
Therefore, (25-^(3/2))^(1/3) evaluates to 2.
2. Evaluate 27^(2/27)^(4/3):
To evaluate this expression, we'll simplify the innermost exponent, which is 2/27.
Let's first calculate 27^(2/27). This means we need to find the 27th root of 27^2. The 27th root of 27^2 is 3.
Now, we'll apply the next exponent, which is 4/3, to the previous result of 3. This means we need to raise 3 to the power of 4/3.
To simplify this further, we can rewrite 4/3 as 1 and 1/3. Raising 3 to the power of 1 gives us 3. Raising 3 to the power of 1/3 means finding the cube root of 3, which is approximately 1.442.
Therefore, 27^(2/27)^(4/3) evaluates to approximately 1.442.
3. Evaluate (16^(5/4) • 16^(1/4))/(16^(1/2))^72:
Let's start by simplifying each part of the expression step by step.
First, let's simplify 16^(5/4). This means we need to find the fourth root of 16^5. The fourth root of 16 is 2, so 16^(5/4) is equal to 2^5, which is 32.
Next, let's simplify 16^(1/4). This means we need to find the fourth root of 16, which is 2.
Now, let's simplify (16^(5/4) • 16^(1/4)). Since both expressions have the same base (16), we can add the exponents together: 5/4 + 1/4 = 6/4, which simplifies to 3/2. So, (16^(5/4) • 16^(1/4)) is equal to 16^(3/2).
Lastly, let's simplify (16^(1/2))^72. This means we need to raise 16^(1/2) to the power of 72. 16^(1/2) is equal to the square root of 16, which is 4. Raising 4 to the power of 72 gives us a large number.
Finally, let's divide (16^(5/4) • 16^(1/4)) by (16^(1/2))^72. Dividing 16^(3/2) by a large number will result in a very small value.
In summary, the expression (16^(5/4) • 16^(1/4))/(16^(1/2))^72 evaluates to a very small value.