What is the answer to 2log_{8}(x+1)+log_{8}(4)=1 ?

The answer to 2log_{8}(x+1)+log_{8}(4)=1 is x=sqrt(2)-1

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2log_{8}(x+1)+log_{8}(4)=1

2 lo g_{8} (x + 1) + lo g_{8} (4) = 1

x = 2 - 1

Decimal Notation

x = 0.41421 \dots

Solution steps

2 lo g_{8} (x + 1) + lo g_{8} (4) = 1

Move

lo g_{8} (4)

to the right side

2 lo g_{8} (x + 1) = 1 - lo g_{8} (4)

Divide both sides by

2

lo g_{8} (x + 1) = \frac{1 - lo g _{8} ( 4 )}{2}

Apply log rules

x + 1 = 2

Solve

x + 1 = 2 : x = 2 - 1

x = 2 - 1

Verify Solutions:

x = 2 - 1

True

The solution is

x = 2 - 1

lo g_{5} (x) = 3^{1}

lo g_{10} (s) = - 5

y, ln (y) = - k t + c

lo g_{10} (X) = - 4.2

15 = lo g_{1.02} (9 x + 1)

What is the answer to 2log_{8}(x+1)+log_{8}(4)=1 ?
The answer to 2log_{8}(x+1)+log_{8}(4)=1 is x=sqrt(2)-1
Solve for x: 2log_{8}(x+1)+log_{8}(4)=1
The solution is x=sqrt(2)-1