I would like to know how much 'n' equals, based on the formula below: $\displaystyle 3^{log_2(n)} = X$
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Originally Posted by mohamedennahdi I would like to know how much 'n' equals, based on the formula below: $\displaystyle 3^{log_2(n)} = X$ Hint: Start by taking log base 3 of both sides. -Dan
What do you think of this? $\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$
Originally Posted by mohamedennahdi What do you think of this? $\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$ You still haven't separated the n from one side. Here's where I was pointing you: $\displaystyle 3^{log_2(n)} = x$ $\displaystyle log_3 \left ( 3^{log_2(n)} \right ) = log_3(x)$ $\displaystyle log_2(n) \cdot log_3(3) = log_3(x)$ Can you finish from here? -Dan
No, please proceed!
Originally Posted by mohamedennahdi No, please proceed! Aw c'mon! Are you telling me you can't solve an equation of the form a*log_2(n) = b? Give it a try! I'll even tell you that log_3(3) = 1. -Dan
How about this: $\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
Originally Posted by mohamedennahdi How about this: $\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$ To get rid of a "log" you have to raise it to a power. So after line 2 we would have $\displaystyle log_2(n) = log_3(x)$ $\displaystyle 2^{log_2(n)} = 2^{log_3(x)}$ $\displaystyle n = 2^{log_3(x)}$ -Dan
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