Thread: indices

Have homework on indices as follows but unable to solve. Any help is appreciated.

Evaluate
1) 12^x/3^x-1

2) 12^x/3^x-1 multiply 4^x+1/2

3) 24^x//2^3x+1 multiply [3^x+1/2^2x+1]^-2

thank you so much

Originally Posted by kristyneo

Have homework on indices as follows but unable to solve. Any help is appreciated.

Evaluate
1) 12^x/3^x-1

2) 12^x/3^x-1 multiply 4^x+1/2

3) 24^x//2^3x+1 multiply [3^x+1/2^2x+1]^-2

thank you so much

Hello,

to #1: I assume that you mean:

$\displaystyle \frac{12^x}{3^{x-1}}=\frac{4^x \cdot 3^x}{\frac13 \cdot 3^x}$ . I'm sure you can continue now

to #2: I assume that you mean:

$\displaystyle \frac{12^x}{3^{x-1}} \cdot 4^{x+\frac12}= \frac{4^x \cdot 3^x \cdot 4^x \cdot 4^{\frac12}}{\frac13 \cdot 3^x}$ . Simplify $\displaystyle 4^{\frac12} = \sqrt{4}=2$ and then continue.

to #3: I can't read the last problem. Please use brackets to differ between bases, exponents, summands and factors.

Thank you for your prompt response.

1) for 1, I cancel out 3x on top and bottom and I got 4x/⅓ which gives
4 x .3

Am i correct?

2) I realise I type wrongly for 2. the question is
(12^x)/3^(x-1) multiply (4^(x+1))/2

3) sorry, typo error again, it should be

24^x/2^(3x+1) multiply [3^(x+1)/2^(2x+1)]^-2

Originally Posted by kristyneo

Thank you for your prompt response.

1) for 1, I cancel out 3x on top and bottom and I got 4x/⅓ which gives
4 x .3

Am i correct?

2) I realise I type wrongly for 2. the question is
(12^x)/3^(x-1) multiply (4^(x+1))/2

3) sorry, typo error again, it should be

24^x/2^(3x+1) multiply [3^(x+1)/2^(2x+1)]^-2

Hello,

I believe that you mean: $\displaystyle 4^x \cdot 3$ which is correct. I would write: $\displaystyle 3 \cdot 4^x$ only to have the constant factor in front.

to #2 (new version):
$\displaystyle \frac{12^x}{3^{x-1}} \cdot \left(4^{x+1}\right)^{\frac12}= \frac{4^x \cdot 3^x \cdot 4^{\frac12 x} \cdot 4^{\frac12}}{\frac13 \cdot 3^x}$ . Consider that $\displaystyle 4^x = (2^2)^x=2^{2x}$ and $\displaystyle 4^{\frac12 x}=(2^2)^{\frac12 x}=2^x$ . Cancel equal terms and collect like terms.

to #3:
$\displaystyle \frac{24^x}{2^{3x+1}}\cdot \left(\frac{3^{x+1}}{2^{2x+1}} \right)^{-2} = \frac{3^x \cdot 8^x}{2 \cdot (2^3)^x}\cdot \left(\frac{2^{2x+1}}{3^{x+1}} \right)^{2} =$ $\displaystyle \frac{3^x \cdot 8^x}{2 \cdot 8^x}\cdot \left(\frac{2^2 \cdot 2^{4x}}{3^2 \cdot 3^{2x}} \right) $ . To continue cancel equal terms and collect like terms.

Hello, kristyneo!

I have to guess what you meant . . .

$\displaystyle 1)\;\frac{12^x}{3^{x-1}}$

We have: .$\displaystyle \frac{(4\cdot3)^x}{3^{x-1}} \;\;=\;\frac{4^x\cdot3^x}{3^{x-1}} \;\;=\;\;4^x\cdot3 \;\;=\;\;3\!\cdot\!4^x$

$\displaystyle 2)\;\frac{12^x}{3^{x-1}} \cdot \frac{4^{x+1}}{2}$

We have: .$\displaystyle \frac{(2^2\cdot3)^x}{3^{x-1}} \cdot \frac{(2^2)^{x+1}}{2}\;\;=\;\;\frac{2^{2x}\cdot3^x }{3^{x-1}} \cdot\frac{2^{2x+2}}{2^1}$

. . $\displaystyle =\;\;\frac{2^{2x}\cdot2^{2x+2}}{2^1}\cdot\frac{3^x }{3^{x-1}} \;\;=\;\;2^{3x+1}\cdot3 \;=\;3\cdot 2^{3x+1}$

$\displaystyle 3)\;\frac{24^x}{2^{3x+1}} \cdot \left[\frac{3^{x+1}}{2^{2x+1}}\right]^{-2} $

We have: .$\displaystyle \frac{(2^3\cdot3)^x}{2^{3x+1}}\cdot \frac{(3^{x+1})^{-2}} {(2^{2x+1})^{-2}} \;=\;\frac{2^{3x}\cdot3^x}{2^{3x+1}} \cdot\frac{3^{-2x-2}}{2^{-4x-2}} \;= \;\frac{2^{3x}\cdot 3^x\cdot3^{-2x-2}}{2^{3x+1}\cdot2^{-4x-2}}$

. . $\displaystyle = \;2^{4x+1}\cdot3^{-x-1} \;=\;2^{4x+1}\cdot3^{-(x+1)} \;=\;\frac{2^{4x+1}}{3^{x+1}}$

Thank you all,
You have been a great help !!!