What is the answer to log_{4}(n+1)-log_{4}(n-2)=1 ?

The answer to log_{4}(n+1)-log_{4}(n-2)=1 is n=3

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log_{4}(n+1)-log_{4}(n-2)=1

lo g_{4} (n + 1) - lo g_{4} (n - 2) = 1

n = 3

Solution steps

lo g_{4} (n + 1) - lo g_{4} (n - 2) = 1

Add

lo g_{4} (n - 2)

to both sides

lo g_{4} (n + 1) - lo g_{4} (n - 2) + lo g_{4} (n - 2) = 1 + lo g_{4} (n - 2)

Simplify

lo g_{4} (n + 1) = 1 + lo g_{4} (n - 2)

Apply log rules

n + 1 = 4 (n - 2)

Solve

n + 1 = 4 (n - 2) : n = 3

n = 3

Verify Solutions:

n = 3

True

The solution is

n = 3

lo g_{4} (11 x) = 3

lo g_{x} (65) = - 6

x, 8 ln (x) = 1

lo g_{x} (\frac{1}{36}) = 2

y, ln (y) = 4 x + 3

What is the answer to log_{4}(n+1)-log_{4}(n-2)=1 ?
The answer to log_{4}(n+1)-log_{4}(n-2)=1 is n=3
Solve for n: log_{4}(n+1)-log_{4}(n-2)=1
The solution is n=3