Logging out soon ( it's still about logs)

• October 13th 2012, 03:45 PM
M670
Logging out soon ( it's still about logs)
Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png: Can I not rewirte this as x=4/7(log6(sqrt108(2048))/(2/sqrt3)^3(9)^3)
then i solve it an get 5.874

I am sure its x=4/7(log6(100077696)
• October 13th 2012, 05:05 PM
Plato
Re: Logging out soon ( it's still about logs)
Quote:

Originally Posted by M670
Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png: Can I not rewirte this as x=4/7(log6(sqrt108(2048))/(2/sqrt3)^3(9)^3)
then i solve it an get 5.874
I am sure its x=4/7(log6(100077696)

Here is a hint.
$\log_6(\sqrt{108})=0.5(\log_6(6^2\cdot 3)=0.5(2+\log_6(3))$
• October 13th 2012, 06:22 PM
Soroban
Re: Logging out soon ( it's still about logs)
Hello, M670!

Quote:

Solve for $x\!:\;x \;=\;\tfrac{4}{7}\left[\log_6\sqrt{108} - 3\log_6\left(\tfrac{2}{\sqrt{3}}\right) + 3\log_6(9) + \log_6(2028)\right]$

Note that:

$\log_6\sqrt{108} \;=\;\log_6 (2^2\cdot3^3)^{\frac{1}{2}} \;=\;\log_6\left(2\cdot3^{\frac{2}{3}}\right)$

$-3\log\left(\tfrac{2}{\sqrt{3}}\right) \;=\;-3\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right) \;=\; -\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right)^3 \;=\;-\log_6\left(\tfrac{2^3}{3^{\frac{3}{2}}}\right)$
. . . . . . . . . $=\;-\left[\log_6(2^3) - \log_6(3^{\frac{3}{2}})\right] \;=\;\log_6(3^{\frac{3}{2}}) - \log_6(2^3)$

$3\log_6(9) \;=\; \log_6(9^3) \;=\; \log_6(3^2)^3 \;=\; \log_6(3^6)$

$\log_6(2048) \;=\;\log_6(2^{11})$

Then we have:

$x \;=\;\tfrac{4}{7}\left[\log_6(2\cdot3^{\frac{3}{2}}) + \log_6(3^{\frac{3}{2}}) - \log_6(2^3) + \log_6(3^6) + \log_6(2^{11})\right]$

$x \;=\;\tfrac{4}{7}\log_6\left[\frac{2\cdot3^{\frac{3}{2}}\cdot3^{\frac{3}{2}}}{2 ^3}\cdot 3^6\cdot 2^{11}\right]$

$x \;=\;\tfrac{4}{7}\log_6(2^9\cdot3^9)$

$x \;=\;\tfrac{4}{7}\log_6(2\cdot3)^9$

$x \;=\;\tfrac{4}{7}\cdot9\log_6(2\cdot3)$

$x \;=\;\tfrac{36}{7}\underbrace{\log_6(6)}_{ \text{This is 1}}$

$x \;=\;\frac{36}{7}$
• October 13th 2012, 08:31 PM
M670
Re: Logging out soon ( it's still about logs)
Quote:

Originally Posted by Soroban
Hello, M670!

Note that:

$\log_6\sqrt{108} \;=\;\log_6 (2^2\cdot3^3)^{\frac{1}{2}} \;=\;\log_6\left(2\cdot3^{\frac{2}{3}}\right)$

$-3\log\left(\tfrac{2}{\sqrt{3}}\right) \;=\;-3\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right) \;=\; -\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right)^3 \;=\;-\log_6\left(\tfrac{2^3}{3^{\frac{3}{2}}}\right)$
. . . . . . . . . $=\;-\left[\log_6(2^3) - \log_6(3^{\frac{3}{2}})\right] \;=\;\log_6(3^{\frac{3}{2}}) - \log_6(2^3)$

$3\log_6(9) \;=\; \log_6(9^3) \;=\; \log_6(3^2)^3 \;=\; \log_6(3^6)$

$\log_6(2048) \;=\;\log_6(2^{11})$

Then we have:

$x \;=\;\tfrac{4}{7}\left[\log_6(2\cdot3^{\frac{3}{2}}) + \log_6(3^{\frac{3}{2}}) - \log_6(2^3) + \log_6(3^6) + \log_6(2^{11})\right]$

$x \;=\;\tfrac{4}{7}\log_6\left[\frac{2\cdot3^{\frac{3}{2}}\cdot3^{\frac{3}{2}}}{2 ^3}\cdot 3^6\cdot 2^{11}\right]$

$x \;=\;\tfrac{4}{7}\log_6(2^9\cdot3^9)$

$x \;=\;\tfrac{4}{7}\log_6(2\cdot3)^9$

$x \;=\;\tfrac{4}{7}\cdot9\log_6(2\cdot3)$

$x \;=\;\tfrac{36}{7}\underbrace{\log_6(6)}_{ \text{This is 1}}$

$x \;=\;\frac{36}{7}$

Wow almost every rule is being used in the problem