I need some help learning to simplify problems like: 125^1/3?

Question

Simplify as much as possible making all exponents positive. Assume all variables represent positive numbers.
125^1/3

12^-3/2

x^3/5 * x^1/2

(9x^4y^2) ^1/2

(27x^3) ^-1/3
*I missed this day of class, can anyone help??

Answer 1

5^3 = 125 then 125^(1/3) = 5

12^-3/2 = 1/sqrt(12^3) = 1/sqrt(1728) = 1/(24sqrt(3)) = sqrt(3)/72

x^3/5 * x^1/2 = x^(3/5+1/2) = x^(11/10)

(9x^4y^2) ^1/2 = sqrt(9)x^(4/2)*y^(2/2) = 3x²y

(27x^3) ^-1/3 = 27^(-1/3) * x^(-3/3) = 1/(3x)

Answer 2

125^1/3 means the cube root of 125
125 = 5 x 5 x 5 = 5^3
Hence the cube root of 125 = 5

12^-3/2 means 1/(12^3/2) NB the negative sign in the 'power' has 'disappeared.
Taking just 12^3/2 = (12^3)^1/2 : This means that 12 is raised to the power of three. When that result is known the square root is calculated.
12^3 = 12 x 12 x 12 = 1728
The sq rt of 1728 is now calculated
1728^1/2 = 41.56921938
Hence 12^3/2 = 41.56921938
Hence 12^-3/2 = 1/41.569219938 = 0.024056261
Or 0.024 (2 sig.fig). The Answer.

x^3/5 * x^1/2.
When multiplying terms with powers , just add the power number.
Hence 3/5 + 1/2 = 11/10 (an improper fraction)
x^3/5 * x^1/2 = x^11/10

(9x^4y^2)^1/2 = 3x^2y^1 = 3x^2y

(27x^3)^-1/3 = 1/(27x^3)^1/3 = 1/(3x)

NB The rules for calculating indices are;_
x^m * x^n = x^(m+n)
x^m / x^n = x^(m-n)
(x^m)^n = x^mn (That is multply the indices together.)
x^-m = 1/x^m (That is the reciprocal).

Hope this helps.!!!!!

Answer 3

125^1/3 means the cube root, or third root of 125, that is what number multiplied by itself 3 times equals 125.

We write 125 as 5^3, since 125 = 5 x 5 x 5, so

125^1/3 = (5^3)^1/3

In general, (m^a)^b = m^ab, so

(5^3)^1/3

= 5^(3 x 1/3)

= 5^1,

= 5.
In general a^b/x means the bth root of a^x.


x^3/5*x ^1/2.

In general, x^a * x^b = x^(a + b), so

x^3/5 * x^1/2

= x^(3/5 + 1/2)

=x^(6/10 + 5/10)

= x^(11/10)

(9x^4y^2)^1/2 means the square root of (9x^4y^2)

The square root of 9x^4 is 3x^2, and the square root of y^2 is y, so we get:

(9x^4y^2)^1/2 = 3x^2y.

In general, N^-p = 1/N^p (In words, N raised to the power of -- p is one over N raised to the power of p), so

(27x^3)^-1/3

= 1/(27^1/3)..........{ 27 = 3^3}, so

= 1/(3^3)^1/3

= 1/3^(3 x 1/3)

= 1/3^1

= 1/3.

Since you have just started this, it would be useful if you commit the following list to memory:

4 = 2^2

8 = 2^3

9 = 3^2

16 = 2^4, and 4^2

25 = 5^2

27 = 3^3

32 = 2^5

36 = 6^2

49 = 7^2

64 = 2^6 and 4^3 and 8^2

81 = 3^2 and 3^4

100 = 10^2

121 = 11^2

125 = 5^3

128 = 2^7

There are obviously lots more, but these will help in the meantime

! ! ! Hope all this lot helps, Twiggy.

Answer 4

Here are some tips that might help. It relates to using the properties of exponents. If an expression that has terms that are multiplying, is raised to an exponent, the exponent can be distributed among them. For example, and expression like ( a. b. c ) ^ d will be equal to (a^d).(b^d).(c^d).
In addition, you will need to remember that exponents multiply each other. For example, take (e ^ f) ^ g it will result in e ^ (f.g). This should help you solve the problem.
In the case of the actual numbers, you will need to find which exponent they could come from ( For example, 25 = 5^; 8 = 2^3....) Good luck!