What is the answer to log_{2}(x+1)+log_{2}(3x-1)=5 ?

The answer to log_{2}(x+1)+log_{2}(3x-1)=5 is x=3

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log_{2}(x+1)+log_{2}(3x-1)=5

lo g_{2} (x + 1) + lo g_{2} (3 x - 1) = 5

x = 3

Solution steps

lo g_{2} (x + 1) + lo g_{2} (3 x - 1) = 5

Apply log rules

(x + 1) (3 x - 1) = 32

Solve

(x + 1) (3 x - 1) = 32 : x = 3, x = - \frac{11}{3}

x = 3, x = - \frac{11}{3}

Verify Solutions:

x = 3

True

, x = - \frac{11}{3}

False

The solution is

x = 3

1.557 = lo g_{10} (\frac{x}{10})

ln (x) + ln (x - 6) = ln (5 x - 28)

lo g_{2} ((2 x + 4)^{3}) = 18

2 lo g_{10} (x + 1) = 5

y, ln (y) = x^{2} + ln (3)

What is the answer to log_{2}(x+1)+log_{2}(3x-1)=5 ?
The answer to log_{2}(x+1)+log_{2}(3x-1)=5 is x=3
Solve for x: log_{2}(x+1)+log_{2}(3x-1)=5
The solution is x=3