# Find value of n

• Jan 7th 2010, 09:30 AM
Mukilab
Find value of n

$\displaystyle 3\cdot \sqrt{27}=3^{n}$
• Jan 7th 2010, 10:03 AM
Dinkydoe
Divide both sides by 3 to obtain: $\displaystyle \sqrt{27} = 3^{n-1}$

Now take the square of both sides to get: $\displaystyle 27 = 3^{2(n-1)}$

We know that $\displaystyle 27 = 3^3 = 3^{2(n-1)}$ so $\displaystyle 3 = 2(n-1)$.

This gives: $\displaystyle 5 = 2n$ so $\displaystyle \frac{5}{2} = n$
(you can ofcourse save some steps: $\displaystyle 3\cdot \sqrt{27} = 3\cdot \sqrt{3^3} = 3^{\frac{3}{2}+1} = 3^{\frac{5}{2}} \rightarrow n = \frac{5}{2}$)
• Jan 7th 2010, 10:37 AM
Mukilab
Quote:

Originally Posted by Dinkydoe
Divide both sides by 3 to obtain: $\displaystyle \sqrt{27} = 3^{n-1}$

Now take the square of both sides to get: $\displaystyle 27 = 3^{2(n-1)}$

We know that $\displaystyle 27 = 3^3 = 3^{2(n-1)}$ so $\displaystyle 3 = 2(n-1)$.

This gives: $\displaystyle 5 = 2n$ so $\displaystyle \frac{5}{2} = n$
(you can ofcourse save some steps: $\displaystyle 3\cdot \sqrt{27} = 3\cdot \sqrt{3^3} = 3^{\frac{3}{2}+1} = 3^{\frac{5}{2}} \rightarrow n = \frac{5}{2}$)

How did you get 3=2(n-1) from $\displaystyle 3^{2(n-1)}??/$
• Jan 7th 2010, 10:43 AM
Dinkydoe
if $\displaystyle 3^3 = 3^{2(n+1)}$ this means that $\displaystyle 3 = 2(n+1)$

Since $\displaystyle a^b = a^c \rightarrow b = c$
• Jan 7th 2010, 12:15 PM
Mukilab
Thanks for the formula I can add to my list!