What is the answer to log_{4}(x+1)+log_{4}(5)=2 ?

The answer to log_{4}(x+1)+log_{4}(5)=2 is x= 11/5

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log_{4}(x+1)+log_{4}(5)=2

lo g_{4} (x + 1) + lo g_{4} (5) = 2

x = \frac{11}{5}

Solution steps

lo g_{4} (x + 1) + lo g_{4} (5) = 2

Move

lo g_{4} (5)

to the right side

lo g_{4} (x + 1) = 2 - lo g_{4} (5)

Apply log rules

x + 1 = \frac{16}{5}

Solve

x + 1 = \frac{16}{5} : x = \frac{11}{5}

x = \frac{11}{5}

Verify Solutions:

x = \frac{11}{5}

True

The solution is

x = \frac{11}{5}

lo g_{x} (\frac{27}{64}) = 3

x^{2} ln (x) - 4 ln (x) = 0

lo g_{5} (4 x - 7) = lo g_{5} (x + 5)

lo g_{10} (x^{64}) = - 3

lo g_{6} (7 x + 1) = 2

What is the answer to log_{4}(x+1)+log_{4}(5)=2 ?
The answer to log_{4}(x+1)+log_{4}(5)=2 is x= 11/5
Solve for x: log_{4}(x+1)+log_{4}(5)=2
The solution is x= 11/5