log8(n-3)+log8(n+4)=1

Question

log8(n-3)+log8(n+4)=1

Answer 1

log₈(n-3) + log₈(n+4) = 1

 

log₈[(n-3)(n+4)] = 1

 

(n-3)(n+4) = 8¹

 

n²+n-20 = 0

 

(n+5)(n-4) = 0

 

n = -5 or 4

 

-5 does not satisfy the original equation (we can't take logarithms of negative numbers)  

 

So, the only solution is 4