# Solve step by step please: Log8 (n-30) + Log8 (n+4) = 1

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log (n-3) base8 +log (n+4) base 8 = 1.

log[(n-30)(n+4)] base8 = 1, as log a+log b = lo ab.

Taking antilogarithm on both sides,

(n-30)(n+4) = 8. As log a to base b = x implies a = b^x.Therefore,

n^2+4n-30n-120= 8. Or

n^2- 26n-128 = 0

n1 ={26+sqrt(26^2-4*128)]/2 = 13+sqrt41 Or n2= 13-sqrt41

log[8](n-30) + log[8](n+4) = 1

By the property of logarithms we can write

log[8](n-30)(n+4) = 1

We know that the log of any number to the base of the same number is 1

therefore we can write,

log[8](n-30)(n+4) = log[8]8

(n-30)(n+4) = 8

n^2 -26n - 120 = 8

n^2- 26n-128 = 0

n ={26+sqrt(26^2-4*128)]/2 = 13+sqrt41 Or n= 13-sqrt41

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