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i also did this one is it correct?
Originally Posted by realistic HI $\displaystyle \log_{3}\sqrt{3^5}=\log_{3}(3^{\frac{5}{2}})$ $\displaystyle =\frac{5}{2}log_{3}3$ $\displaystyle =\frac{5}{2} $
Originally Posted by realistic i also did this one is it correct? yes its correct.
Originally Posted by mathaddict HI $\displaystyle \log_{3}\sqrt{3^5}=\log_{3}(3^{\frac{5}{3}})$ $\displaystyle =\frac{5}{3}log_{3}3$ $\displaystyle =\frac{5}{3} $ in the back of the book they have $\displaystyle =\frac{5}{2} $ is that even possible for this question?
Originally Posted by realistic in the back of the book they have $\displaystyle =\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 .
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
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