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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) = 14

I'm sure, that n isn't natural number, because:

4^1 + 4^(-1) = 4 + (1/4)^1 = 4 + 1/4 = 4.25

and

4^2 + 4^(-2) = 16 + (1/4)^2 = 16 + 1/16 = 16.0625

14 is greater than 4.25 and is lower than 16.0625, so:

1 < n < 2

I don't have any idea what's next. Can you help me?


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

x + 1/x = 14 or

x2 - 14x + 1 = 0

which is just a quadratic equation in x. Once you found it,

n = logx / log4 = log4x = (log2x)/2.


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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 :)


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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) / 2

4^n = 7 +/- 48^0.5

then by logging both sides

n log 4 = log (7 +/- 48^0.5)

n = log (7 +/- 48^0.5) / log 4

n = 1.89997 , -1.89997


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