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What are the potential solutions of log4x+log4 (x+6)=2
x=-2 and x=-8
x=-2 and x=8
x=2 and x=-8
x=2 and x=8

Sagot :

fieldscr

Answer:

log4x+log4(x-6)=2

place under single log using multiplication rule

log4(x(x-6))=2

convert to exponential form:(base(4) raised to log of number(2)=number(x(x-6)

4^2=x(x-6)

16=x^2-6x

x^2-6x-16=0

(x-8)(x+2)=0

x=8

or

x=-2 (reject, x>0)

Step-by-step explanation:

fieryanswererft

Answer:

[tex]\sf x=2, \ x = -8[/tex]

solving steps:

[tex]\hookrightarrow \sf log_4 x+log_4 (x+6)=2[/tex]

[tex]\hookrightarrow \sf log_4 (x(x+6))=log_4 (16)[/tex]

[tex]\hookrightarrow \sf log_4 (x^2+6x)=log_4 (16)[/tex]

[tex]\hookrightarrow \sf x^2+6x=16[/tex]

[tex]\hookrightarrow \sf x^2+6x-16=0[/tex]

[tex]\hookrightarrow \sf x^2+8x-2x-16=0[/tex]

[tex]\hookrightarrow \sf x(x+8)-2(x+8)=0[/tex]

[tex]\hookrightarrow \sf (x-2)(x+8)=0[/tex]

[tex]\hookrightarrow \sf x=2, \ x = -8[/tex]

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