Respuesta :

AnthrαX
[tex]-log(x+8)=4-log(x-7)[/tex] 

[tex]-\log _{10}\left(x+8\right)+\log _{10}\left(x+8\right)=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)[/tex] 

[tex]0=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)[/tex] 

[tex]0+\log _{10}\left(x-7\right)=4-\log _{10}\left(x-7\right)+\log _{10}\left(x+8\right)+\log _{10}\left(x-7\right)[/tex] 

[tex]\log _{10}\left(x-7\right)=\log _{10}\left(x+8\right)+4[/tex] 

[tex]\log _{10}\left(x-7\right)=\log _{10}\left(x+8\right)+\log _{10}\left(10000\right)[/tex] 

[tex]x-7=\left(x+8\right)\cdot \:10000[/tex] 

[tex]\mathrm{Solve\:}\:x-7=\left(x+8\right)10000:\quad x=-\dfrac{26669}{3333}[/tex] 

[tex]\mathrm{Verifying\:Solutions}:\quad x=-\dfrac{26669}{3333}\space\mathrm{False}[/tex] 

[tex]\mathrm{No\:Solution\:for\:x\in \mathbb{R}}[/tex]
altavistard
Note that -log(x+8) + log(x-7) = 4, and that the left side is equal to

            x-7
log ------------- 
            x-8

Therefore, 

            x-7
log ------------- = 4
            x-8

Acknowledging that your "log" actually represents "log to the base 2 of ... "

We get:

   x-7
--------- = 2^4 = 16
   x-8

Can this be solved for x?  

Rearranging,         x-7 = 16x - 128, or -7 = 15x - 128, or 121 = 15x
                                                    121
Dividing 121 by 15, we get   x = -------  = 121/15 = approx. 8.067.
                                                      15

                                                      
So far I see no reason why the given -log(x+8)=4-log(x-7) "has no solution."

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