What is the answer to log_{2}(t)+log_{2}(t+1)=1 ?

The answer to log_{2}(t)+log_{2}(t+1)=1 is t=1

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log_{2}(t)+log_{2}(t+1)=1

lo g_{2} (t) + lo g_{2} (t + 1) = 1

t = 1

Solution steps

lo g_{2} (t) + lo g_{2} (t + 1) = 1

Apply log rules

t (t + 1) = 2

Solve

t (t + 1) = 2 : t = 1, t = - 2

t = 1, t = - 2

Verify Solutions:

t = 1

True

, t = - 2

False

The solution is

t = 1

ln ((3 x)^{2}) = 16

ln (x + 25) = ln (x) + ln (25)

lo g_{3} (x + 5) - lo g_{3} (x - 1) = 1

lo g_{10} (x) = 2.68

lo g_{2} (6 x + 70) = 6

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