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There are several real numbers \(x\) such that \(log_2 (x + 2), \ \log_4 (3x + 4), \ \log_8 (7x + 8)\)
are three real numbers in arithmetic progression in the order listed. One such number \(x\) can be expressed in the form \(\frac{-a + b \sqrt{c}}{d},\)
where  \(a,b,c,d\)   are positive integers. Find  \(a+b+c+d\)(It is assumed that the number in this form is as simplified as possible.)

 Jun 22, 2019
 #1
avatar+104688 
+2

Using the change-of base theorem  we have

 

log (3x + 4)             log (3x + 4)         log (3x + 4)               (1/2)log(3x + 4)

_________    =      __________  =   ______________=    ____________    (1)

log 4                          log 2^2                2 log 2                         log 2

 

log ( 7x + 8)        log(7x + 8)           log (7x + 8)                (1/3)log(7x + 8)

_________   =  __________  =  _____________ =        _____________   (2)

log 8                    log 2^3                  3 log 2                           log 2

 

The arithmetic difference must  be   (2) - (1)  =

    

(1/3) log (7x + 8) - (1/2)log(3x + 4)

__________________________

                  log 2

 

This implies that

 

log (x + 2)               (1/3)log(7x + 8) - (1/2)log (3x + 4)              (1/2)log(3x + 4)

________    +        ____________________________   =    ______________

  log   2                                  log 2                                                log 2

 

 

 

;log (x + 2)  +  (1/3)log(7x + 8) - (1/2)log(3x + 4)  =  (1/2)log(3x + 4)

 

log (x + 2) + (1/3)log(7x + 8)    =   log(3x + 4)

 

3 log( x + 2)  + log(7x + 8)  = 3log(3x + 4)

 

log ( x + 2)^3 + log(7x + 8)    =  log (3x + 4)^3

 

log  [ (x + 2)^3 * (7x + 8)]  = log(3x + 4)^3

 

Which implies that

 

(x + 2)^3 (7x + 8)  = (3x + 4)^3

 

(x^3 + 3x^2*2 + 3x*4 + 2^3) (7x + 8)   =  (3x)^3  + 3* (3x)^2*4 + 3* 3x*4^2 + 4^3

 

( x^3 + 6x^2 + 12x + 8)(7x + 8)  =  27x^3 + 108x^2 + 144x + 64

 

7 x^4 + 50 x^3 + 132 x^2 + 152 x + 64  =  27x^3 + 108x^2 + 144x + 64

 

7x^4 + 23x^3 + 24x^2 + 8x  = 0

 

x (7x^3 + 23x^2 + 24x + 8)  = 0

 

x = 0 is one solution

 

7x^3 + 23x^2 + 24x + 8  = 0

 

Using the Rational zeroes theorem, I see that x = -1 is also another solution

 

So....we can use synthetic division to find the residual polynomial

 

 

-1  [  7     23     24      8   ]

                -7     -16    -8

       ________________

        7     16       8       0

 

 

The   remaining polynomial is     7x^2  + 16x + 8

 

Using the Quadratic Formula  

 

x =    -16 ±√ [ 16^2  - 4(7)(8) ]      =          -16  ±√ 32      =      -16 ± 4√2    =     -8 ±2√2

        ____________________                __________          _________        _______

                   2 *7                                            14                          14                       7

 

The solution  -8 - 2√2

                     _______  ≈  -1.55       which makes two of the original logs undefined  

                           7

 

So

 

 

x  =   -8 + 2 √2

        _________  ≈  -.74

               7

 

a =  8   b =  2    c  = 2   d  = 7

 

And their sum  =    19

 

 

cool cool cool     

 Jun 22, 2019
 #2
avatar+166 
0

Thank you

 Jun 29, 2019

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