2/the 6th root of 81
the answer gave an answer of "2 3rd root of 3/3"
sorry i didn't type the root signs, i dunno how.
can you show me how to work this?
the book gave me that answer.
2/the 6th root of 81
= 2/[(3^4)^(1/6)]
= 2/3^(2/3)
now multiply top and bottom by 3^(1/3)
which would give you 3^(3/3) or 3 in the bottom
= 2(3^(1/3))/3
= 2*cuberoot(3)/3, your answer.
Of course! I can help you understand how to work out the expression: 2 divided by the 6th root of 81.
To start, let's first find the value of the 6th root of 81. The word "root" refers to finding a number that, when raised to a certain power, gives 81. In this case, we are looking for the number that, when raised to the power of 6, gives us 81.
To find the 6th root of 81, you can raise 81 to the power of 1/6. Mathematically, it is represented as follows:
6th root of 81 = 81^(1/6).
To simplify this expression, you can rewrite 81 as a power of 3 using the fact that 81 = 3^4:
81^(1/6) = (3^4)^(1/6).
When raising a power to another power, you multiply the exponents, so we have:
(3^4)^(1/6) = 3^(4*(1/6)).
Simplifying further, we find:
3^(4*(1/6)) = 3^(4/6).
The exponent 4/6 can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2. After simplification, we have:
3^(4/6) = 3^(2/3).
So, the 6th root of 81 is equivalent to 3^(2/3).
Now, let's solve the expression 2 divided by the 6th root of 81:
2 / (6th root of 81) = 2 / (3^(2/3)).
To divide by a number with a fractional exponent, you can rewrite it using negative exponents. In this case, since 2/3 is the exponent, we can rewrite it as follows:
2 / (3^(2/3)) = 2 / (3^(2*1/3)) = 2 / (3^(2*(1/3))) = 2 / (3^(2/3)).
Using the rule of exponents, when dividing with the same base but different exponents, we subtract the exponents:
2 / (3^(2/3)) = 2 / 3^2/3 = 2 / 3^(2/3) = 2 * 3^-(2/3).
Therefore, the final simplified answer is:
2 / (6th root of 81) = 2 * 3^-(2/3).
Please note that the root signs were not included in the final answer because they were not specified in the question.