how to re-arrange this

**realistic** · November 13th 2009, 06:04 PM

i also did this one is it correct?

**mathaddict** · November 13th 2009, 06:12 PM

Originally Posted by realistic

HI

$\log_{3}\sqrt{3^5}=\log_{3}(3^{\frac{5}{2}})$

$=\frac{5}{2}log_{3}3$

$ =\frac{5}{2} $

**mathaddict** · November 13th 2009, 06:13 PM

Originally Posted by realistic

i also did this one is it correct?

yes its correct.

**realistic** · November 13th 2009, 06:14 PM

Originally Posted by mathaddict

HI

$\log_{3}\sqrt{3^5}=\log_{3}(3^{\frac{5}{3}})$

$=\frac{5}{3}log_{3}3$

$ =\frac{5}{3} $

in the back of the book they have $ =\frac{5}{2} $

is that even possible for this question?

**mathaddict** · November 13th 2009, 06:16 PM

Originally Posted by realistic

in the back of the book they have $ =\frac{5}{2} $

is that even possible for this question?

oops , the book is correct . i misread the 3 as cube root . By the way , i edited mypost .

**realistic** · November 13th 2009, 06:27 PM

Originally Posted by mathaddict

oops , the book is correct . i misread the 3 as cube root . By the way , i edited mypost .

ok thanks i c how it is done now

Thread: how to re-arrange this