Solve for x.

log (base 7) 5x-1 = log (base 7) x+7

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`log_7 (5x-1)=log_7 (x+7)`

`log_7( 5x-1)-log_7(x+7)=0`

`because`

`log_7 x-log_7 y=log_7(x/y)`

`therefore`

`log_7((5x-1)/(x+7))=0`

`log_7 1=0`

`therefore`

`(5x-1)/(x+7)=1`

`5x-1=(x+7)xx1`

`5x-x=7+1`

`4x=8`

`x=2`

Thus answer is

X=2

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