Logarithims
- Thread starter bettyh
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,325
This is wonderful problem. i don't want to spoil it for you.
\(\displaystyle log_{16}(27) = \frac{log(27)}{log(16)} = \frac{log(3^{3})}{log(4^{2})} = \frac{log(3^{3})}{2\cdot log(4)} = \frac{(1/2)\cdot log(3^{3})}{log(4)} = \frac{log(3^{3/2})}{log(4)} = log_{4}(3^{3/2})\)
\(\displaystyle 4^{log_{16}(27)} = 4^{log_{4}(3^{3/2})} = 3^{3/2}\)
Now what?
\(\displaystyle log_{16}(27) = \frac{log(27)}{log(16)} = \frac{log(3^{3})}{log(4^{2})} = \frac{log(3^{3})}{2\cdot log(4)} = \frac{(1/2)\cdot log(3^{3})}{log(4)} = \frac{log(3^{3/2})}{log(4)} = log_{4}(3^{3/2})\)
\(\displaystyle 4^{log_{16}(27)} = 4^{log_{4}(3^{3/2})} = 3^{3/2}\)
Now what?