Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Formulas » Logarithm Formulas » 2009-11-25 23:06:46

Hi Shekhar,
Here is the solution.
   ((2/3)^k)n = 1
⇒(2/3)^k = 1/n
⇒(2/3)^k = n-¹
Applying log on both sides.
⇒k log(2/3) = -log n
Multiplying (-) on both sides
⇒k log(2/3)-¹ = log n
⇒k log(3/2) = log n
∴ k = (log n)/(log (3/2))
⇒ k = log n base (3/2) 


Shekhar wrote:

Can anyone prove the following....
Given,
((2/3)^k)n = 1
Required to prove,
k is equal to log of n on base 3/2

Board footer

Powered by FluxBB