# Logs

• October 8th 2012, 06:44 AM
M670
Logs
Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png :http://webwork.mathstat.concordia.ca...d000b7e941.png

log2x9/2=-3
2-3=x9/2
.125x-4.5=x
-0.5625=x

but I don't think it's right
• October 8th 2012, 06:52 AM
skeeter
Re: Logs
$\frac{9}{2} \log_2{x} = -3$

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

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

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

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

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: $\left(x^{9/2}\right)^{2/9}= x= (1/2^3)^{2/9}= 1/2^{6/9}= 1/2^{2/3}= \frac{1}{\sqrt[3]{4}}$

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

Quote:

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

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

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

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

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

but I don't think it's right

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

Raise both sides to the $\tfrac{2}{9}$ power:
. . . $\left(x^{\frac{9}{2}}\right)^{\frac{2}{9}} \:=\:\left(2^{-3}\right)^{\frac{2}{9}} \quad\rightarrow\quad x \:=\:2^{-\frac{2}{3}}$

Therefore: . $x\;=\;0.629960515$
• October 8th 2012, 04:19 PM
M670
Re: 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