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+11
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 Mar 21, 2020
 #1
avatar+111329 
+1

2log2 ( x + 5)  =  log2(x - 9) + log2(x + 53)  +  1

 

Note  that  1  =  log2 2

 

So  we  have

 

2log2 ( x + 5)  =  log2(x - 9) + log2(x + 53)  +   log2 2

 

By a couple of log properties we  can write

 

log2 ( x + 5)^2  =  log[ (x - 9)(x + 53)(2) ]

 

The log base is the same so we can solve  this

 

(x + 5)^2  = (x - 9) ( x + 53)(2)           simpify

 

x^2 + 10x + 25  =  (x^2  + 53x - 9x  - 477) (2)

 

x^2  + 10 + 25  =  (x^2  + 44x  - 477) (2)

 

x^2  + 10x +  25   = 2x^2  + 88x  - 954    rearrange as

 

x^2 +  78x  - 979   =  0        this can  be  factored as

 

(x  -11) ( x + 89)   =   0

 

Setting both factors to 0  and  solving for  x   gives  us

 

x  = 11    or  x   = -89

 

The second answer gives us the log of a negative  number in the original equation....so..... it's no good

 

The only answer is

 

x  = 11

 

cool cool cool

 Mar 21, 2020
 #2
avatar+1968 
+2

Nice job, Chris!!!

CalTheGreat  Mar 21, 2020
 #3
avatar+111329 
0

THX, CTG   !!!

 

 

 

cool cool cool

CPhill  Mar 21, 2020

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