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 Subject: "raising 4 to the n-th power" Previous Topic | Next Topic
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coolman guest
Oct-29-06, 06:27 AM (EST)

"raising 4 to the n-th power"

 Hello. My friend asked me to slove a math problem, but i can't slove it too:Find all values of 'n' in this equation: 4^n + 4^(-n) = 14I'm sure, that n isn't natural number, because:4^1 + 4^(-1) = 4 + (1/4)^1 = 4 + 1/4 = 4.25and4^2 + 4^(-2) = 16 + (1/4)^2 = 16 + 1/16 = 16.062514 is greater than 4.25 and is lower than 16.0625, so: 1 < n < 2I don't have any idea what's next. Can you help me?

alexb
Charter Member
1919 posts
Oct-29-06, 06:32 AM (EST)    1. "RE: raising 4 to the n-th power"
In response to message #0

 If x = 4n then the equation isx + 1/x = 14 orx2 - 14x + 1 = 0which is just a quadratic equation in x. Once you found it,n = logx / log4 = log4x = (log2x)/2. coolman guest
Oct-29-06, 04:04 PM (EST)

2. "RE: raising 4 to the n-th power"
In response to message #1

 Thanks for help :)

NAL guest
Nov-20-06, 07:31 AM (EST)

3. "RE: raising 4 to the n-th power"
In response to message #0

 If you multiply by 4^n and rearrange you get a quadratic in 4^n -4^n + 4^(-n) = 14(4^n)^2 + 1 = 14*4^n (4^n)^2 - 14*4^n + 1 = 0 so using the quadratic formula -4^n = (14 +/- (14^2 - 4)^0.5) / 24^n = 7 +/- 48^0.5then by logging both sidesn log 4 = log (7 +/- 48^0.5)n = log (7 +/- 48^0.5) / log 4n = 1.89997 , -1.89997

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