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1) Find a/b when \(2\log{(a -2b)} = \log{a} + \log{b}.\)

2) For each positive integer p, let b(p) denote the unique positive integer k such that \(|k-\sqrt{p}|<\frac{1}{2}\). For example, b(6)=2 and b(23)=5. Find \(S=\sum_{p=1}^{2007} b(p)\)

3) William Sydney Porter tried to perform the calculation \(\frac{-3+4i}{1+2i}\). However, he accidentally missed the minus sign, finding \(\frac{3+4i}{1+2i}=\frac{11}{5}-\frac{2}{5}i\). What answer should he have obtained?

 Jul 14, 2019
 #1
avatar+111393 
+1

1.

 

2log ( a - 2b)  = log a + log  b

 

log(a - 2b)^2  = ;og (ab)      which implies that

 

(a - 2b)^2  = ab

 

a^2 - 4ab  + 4b^2  =  ab

 

a^2 - 5ab + 4b^2  =  0    factor

 

(a - 4b)  ( a - b)  = 0

 

So...either   a - 4b  = 0  ⇒   a = 4b   ⇒  a / b   = 4    (1)

 

or

 

a - b  =  0   ⇒  a  = b

If a is positive.....the original log on the left hand side is undefined

If a is negative......log a is undefined..so...

 

a / b   = 4

 

 

 

cool cool cool

 Jul 14, 2019
 #2
avatar
+1

2)    sumfor(n, 1, 44, 2*n^2) + 27*45 = 59,955

 Jul 14, 2019
 #3
avatar+111393 
+1

3)

 

-3 + 4i           [ 1 - 2i ]          -3 + 4i +6i - 8i^2          -3 + 10i - 8(-1)          5 + 10i

_______                       =   ______________  =   _____________  =    ______  =

1  + 2i           [ 1 -2i ]           1 -  4i^2                        1 - 4(-1)                        5

 

 

1 + 2i

 

BTW ....William Sydney Porter  ....better known as  .....  "O Henry "

 

 

 

cool cool cool

 Jul 14, 2019

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