exponents

**Dean** · July 31st 2007, 08:50 PM

Could someone help me out with these?

Using the laws of indices to simplify each of the following, giving answers with positive indices only.

1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)

2. 21^(3x+1).7(2-x)/49^(x-1).27^(x-4)

**tukeywilliams** · July 31st 2007, 11:05 PM

Originally Posted by Dean

Could someone help me out with these?

Using the laws of indices to simplify each of the following, giving answers with positive indices only.

1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)

2. 21^(3x+1).7(2-x)/49^(x-1).27^(x-4)

1. $\frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}} = \frac{3^{3x+4} \cdot 3^{-x-2} \cdot 3^{-x-2} \cdot 2^{-x-2}}{2^{x+3} \cdot 2^{x+3} \cdot 3^{x+3} \cdot 2^{x+3} \cdot 2^{-2x-1} \cdot 2^{-2x-1}} = \frac{3^{x}}{2^{9} \cdot 3^{x+3}} = \frac{1}{3^{3} \cdot 2^{9}}$

2. $\frac{21^{3x+1} \cdot 7^{2-x}}{49^{x-1} \cdot 27^{x-4}} = \frac{7^{3x+1} \cdot 3^{3x+1} \cdot 7^{2-x}}{7^{x-1} \cdot 7^{x-1} \cdot 27^{x-4}} = \frac{7^{2x+3} \cdot 3^{3x+1}}{7^{2x-2} \cdot 3^{x-4} \cdot 3^{x-4} \cdot 3^{x-4}} = \frac{7^{5} \cdot 3^{2x+5}}{3^{2x-8}} = 7^{5} \cdot 3^{13}$

**earboth** · July 31st 2007, 11:18 PM

Originally Posted by Dean

Could someone help me out with these?
Using the laws of indices to simplify each of the following, giving answers with positive indices only.

1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)
...

Hello,

1. I assume that you mean exponents and not indices (?)
2. I assume that you are familiar with the laws and rules to calculate powers.
3. I assume that you want to simplify

$\frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}}$

Factor all bases and transform to a produkt:

$\frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}}$ = $3^{3x} \cdot 3^4 \cdot 3^{-2x} \cdot 2^{-x} \cdot 3^{-4} \cdot 2^{-2} \cdot 3^{-x} \cdot 2^{-3x} \cdot 3^{-3} \cdot 2^{-9} \cdot 2^{4x} \cdot 2^2$ = $3^0 \cdot 2^0 \cdot 3^{-3} \cdot 2^{-9} = \frac{1}{24^3}$

**Dean** · August 1st 2007, 12:07 AM

Which way is correct?

**tukeywilliams** · August 1st 2007, 12:09 AM

$\frac{1}{24^3} = \frac{1}{3^{3} \cdot 2^{9}}$ . Both ways are correct. I just converted the $18^{-x-2}$ to $(3 \cdot 3 \cdot 2)^{-x-2}$ and worked from there. Same with the second problem.

**Dean** · August 1st 2007, 12:40 AM

Oh, thanks so much guys.

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