Indices questions - step by step pls?

Question

a) 12a^7/8 / 3a^1/4
b) (a^2/3 b^5/6)(a^1/2 b) / (ab)^1/3
c) (a^1/4 b^3)^1/2 / (a^1/3 b^1/4)^-5
d) (a^1/2 + a^1/2)(a^1/2 - a^1/2)

Answer 1

Hi,

Here are the rules to follow:

When one term in a parentheses is raised to a power outside the parentheses, multiply that outside exponent times every exponent inside the parentheses. If there is a variable or a number inside the parentheses with no exponent shown, its exponent is a one. If the exponent is negative, move that variable or number into the denominator and make the exponent positive.

In division problems, divide or reduce the coefficients. When a variable is on both top and bottom, subtract the top exponent minus the bottom exponent. Theses variables are now combined in the numerator. If the exponent is negative, move that variable into the denominator and make the exponent positive.

When multiplying in problems, multiply the coefficients together. When the same variable occurs more than once, keep the variable and add its exponents together. If the exponent is negative, move that variable into the denominator and make the exponent positive.

a) 12a^7/8
--------------- This is division. Divide coefficients. Subtract the
..... 3a^1/4...exponents.

Since 7/8 - 1/4 = 7/8 - 2/8 = 5/8, the answer is 4a^(5/8).



b) (a^2/3 b^5/6)(a^1/2 b)
----------------------------------
....... (ab)^1/3

Start by multiplying together the numerator.
On "a",2/3 + 1/2 = 4/6 + 3/6 = 7/6
On "b", 5/6 + 1 = 5/6 + 6/6 = 11/6
The numerator is now a^(7/6)b^(11/6).

In the denominator multiply 1/3 times each exponent of 1 inside the parentheses to get a^(1/3)b^(1/3).

The problem is now:

a^(7/6)b^(11/6)
---------------------
a^(1/3)b^(1/3)

Now in this division, subtract exponents on each variable.

On "a", 7/6 - 1/3 - 7/6 - 2/6 = 5/6.
On "b", 11/6 - 1/3 = 11/6 - 2/6 = 9/6 = 3/2

This simplifies to a^(5/6)b^(3/2) as the answer.


c) (a^1/4 b^3)^1/2 / (a^1/3 b^1/4)^-5

For the first parentheses, multiply 1/2 times each exponent inside the parentheses to get:
a^(1/8)b^(3/2)

For the second parentheses, multiply -5 times each exponent inside the parentheses to get:
a^(-5/3)b^(-5/4)

This division problem is now:

a^(1/8)b^(3/2)
----------------------
a^(-5/3)b^(-5/4)

Now in this division, subtract exponents on each variable.

On "a", 1/8 - (-5/3)= 1/8 + 5/3 = 3/24 + 40/24 = 43/24
On "b", 3/2 - (-5/4) = 3/2 + 5/4 = 6/4 + 5/4 = 11/4

This simplifies to a^(43/24)b^(11/4) as the answer.

d) (a^1/2 + a^1/2)(a^1/2 - a^1/2)

Combine like terms by adding their coefficients.
a^1/2 + a^1/2 = 2a^1/2
(a^1/2 - a^1/2) = 0

Multiplying these together, 2a^1/2 * 0 = 0 <== answer

I hope these help!! :-)

Answer 2

c)

Step by step:

{[a^(1/4) * b^3)]^1/2} / [a^(1/3) * b^(1/4)]^(-5)
= {a^[(1/4)*(1/2)] * b^[3*(1/2)]} / [a^(1/3) * b^(1/4)]^(-5)
= {a^(1/8) * b^[3*(1/2)]} / [a^(1/3) * b^(1/4)]^(-5)
= [a^(1/8) * b^(3/2)] / [a^(1/3) * b^(1/4)]^(-5)
= [a^(1/8) * b^(3/2)] / {a^[(1/3)*(-5)] * b^[(1/4)*(-5)]}
= [a^(1/8) * b^(3/2)] / {a^(-5/3) * b^[(1/4)*(-5)]}
= [a^(1/8) * b^(3/2)] / [a^(-5/3) * b^(-5/4)]
= [a^(1/8) * b^(3/2)] * [a^(5/3) * b^(5/4)]
= a^(1/8) * b^(3/2) * a^(5/3) * b^(5/4)
= a^(1/8) * a^(5/3) * b^(3/2) * b^(5/4)
= a^(5/24) * b^(3/2) * b^(5/4)
= a^(5/24) * b^(15/8)
= a^(5/24) * b * b^(7/8)

Ask if some of the steps are not clear.
-

Answer 3

a) (60*10^8)-(one million.5*10^8)=fifty 8.5*10^8 = 585*10^7 b)(one million/3)^x=one million or one million/(3^x)=one million, no rely, x=0 , notwithstanding selection interior the zeroth ability supplies one million c)3^(x+7)=one million, x+7=0, x=-7 d)((3^2x)*(3^4))/(3^6y) = (3^(2x+4)) / (3^6y)= 3^(2x+4-6y), carry on with the ability policies e) (root5 / (x-5))=one million/5^3, x-5 = 5^(one million/2)*5^3 =5^(7/2), x=5^(7/2)+5 i will not be able to understand the logarithms in question d, there isn't any dazzling section interior the equation, so do not hear the precise Contributor..... reliable success....