What is the answer to 3=log_{3}(t+3)+log_{3}(t-3) ?

The answer to 3=log_{3}(t+3)+log_{3}(t-3) is t=6

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3=log_{3}(t+3)+log_{3}(t-3)

3 = lo g_{3} (t + 3) + lo g_{3} (t - 3)

t = 6

Solution steps

3 = lo g_{3} (t + 3) + lo g_{3} (t - 3)

Apply log rules

27 = (t + 3) (t - 3)

Solve

27 = (t + 3) (t - 3) : t = 6, t = - 6

t = 6, t = - 6

Verify Solutions:

t = 6

True

, t = - 6

False

The solution is

t = 6

lo g_{25} (x + 2) = lo g_{5} (x)

lo g_{5} (t^{2} + 6 t) = lo g_{5} (7)

ln (4 t) - ln (3 t) = 2

lo g_{10} (m) = - 1 + lo g_{10} (m + 3)

lo g_{2} (4 x + 5) - lo g_{2} (4 - x) = 3

What is the answer to 3=log_{3}(t+3)+log_{3}(t-3) ?
The answer to 3=log_{3}(t+3)+log_{3}(t-3) is t=6
Solve for t: 3=log_{3}(t+3)+log_{3}(t-3)
The solution is t=6