Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png :http://webwork.mathstat.concordia.ca...d000b7e941.png

I made it into

log_{2}x^{9/2}=-3

2^{-3}=x^{9/2 .125x-4.5=x -0.5625=x but I don't think it's right}

Printable View

- Oct 8th 2012, 06:44 AMM670Logs
Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png :http://webwork.mathstat.concordia.ca...d000b7e941.png

I made it into

log_{2}x^{9/2}=-3

2^{-3}=x^{9/2 .125x-4.5=x -0.5625=x but I don't think it's right} - Oct 8th 2012, 06:52 AMskeeterRe: Logs
$\displaystyle \frac{9}{2} \log_2{x} = -3$

$\displaystyle \log_2{x} = -\frac{2}{3}$

$\displaystyle x = 2^{-\frac{2}{3}} = \frac{1}{\sqrt[3]{4}}$ - Oct 8th 2012, 07:09 AMHallsofIvyRe: Logs
Skeeter's method, dividing both sides by 9/2 first, is simpler than yours but it can be done your way:

$\displaystyle \frac{9}{2}log_2(x)= -3$

$\displaystyle log_2(x^{9/2})= -3$

$\displaystyle x^{9/2}= 2^{-3}= \frac{1}{8}$

I don't know how you now got ".125x-4.5=x". Was that first "x" supposed to be a multiplication sign? Bad idea! Never use the same symbol to mean two different things!

But even so, you get rid of the "9/2" power on the left by taking the 9/2 root- which is the same thing as the 2/9 power, on both sides: $\displaystyle \left(x^{9/2}\right)^{2/9}= x= (1/2^3)^{2/9}= 1/2^{6/9}= 1/2^{2/3}= \frac{1}{\sqrt[3]{4}}$

If you want or are required to "rationalize the denominator", multiply both numerator and denominator by $\displaystyle \left(\sqrt[3]{4}\right)^2= \sqrt[3]{16}$: $\displaystyle \frac{\sqrt[3]{16}}{4}$. - Oct 8th 2012, 07:12 AMSorobanRe: Logs
Hello, M670!

Quote:

Solve for $\displaystyle x\!:\;\tfrac{9}{2}\log_2x \:=\:-3$

I made it into:

$\displaystyle \log_2\left(x^{\frac{9}{2}}\right) \:=\:-3$ . . Right!

$\displaystyle 2^{-3} \:=\:x^{\frac{9}{2}}$ . . Correct!

$\displaystyle 0.125x-4.5\:=\:x$ . . I don't understand this step.

$\displaystyle -0.5625\:=\:x$ . . Nor this step.

but I don't think it's right

In your second step, you had: $\displaystyle x^{\frac{9}{2}} \:=\:2^{-3}$

Raise both sides to the $\displaystyle \tfrac{2}{9}$ power:

. . . $\displaystyle \left(x^{\frac{9}{2}}\right)^{\frac{2}{9}} \:=\:\left(2^{-3}\right)^{\frac{2}{9}} \quad\rightarrow\quad x \:=\:2^{-\frac{2}{3}} $

Therefore: .$\displaystyle x\;=\;0.629960515$

- Oct 8th 2012, 04:19 PMM670Re: Logs
sorry guys I should of written my step .125x-4.5=x as follows .125(-4.5)=x

but i understand what you have all shown me