Solving equation with logs

**metaname90** · September 12th 2010, 10:04 AM

With the help of technology I've figured out that r=27 but I can't quite work it out.

I know 2^3log4(3)=2^log4(27 or 2^(log27)/(log4) and sqrt(r)=r^(1/2) but I can't quite put it all together....

**Plato** · September 12th 2010, 10:23 AM

First notice that $\log_4(2)=\frac{1}{2}$
Then $\displaystyle 2^{3\log_4(3)}= 2^{\log_4(27)}$ .
Take the log of both $\log_4(27)\log_4(2)= \frac{1}{2}\log_4(27)= \log_4(\sqrt{27})=\log_4(\sqrt{r})$

**Soroban** · September 12th 2010, 01:30 PM

Hello, metaname90!

$\text{Solve for }r\!:\;\;2^{3\log_4(3)} \:=\:\sqrt{r}$

The left side is: . $2^{\log_4(3^3)} \;=\;2^{\log_4(27)}$

Since $2 = 4^{\frac{1}{2}}$
. . we have: . $(4^{\frac{1}{2}})^{\log_4(27)} \;=\; 4^{\frac{1}{2}\log_4(27)} \;=\;4^{\log_4(27^{\frac{1}{2}})} \;=\;27^{\frac{1}{2}}$

The equation becomes: . $27^{\frac{1}{2}} \:=\:r^{\frac{1}{2}} \quad\Rightarrow\quad r \:=\:27$

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