What is the answer to 2x-(log_{3}(4)+x+1)+3=0 ?

The answer to 2x-(log_{3}(4)+x+1)+3=0 is x=2log_{3}(2)-2

Solve for x: 2x-(log_{3}(4)+x+1)+3=0

The solution is x=2log_{3}(2)-2

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2 x - (lo g_{3} (4) + x + 1) + 3 = 0

x = 2 lo g_{3} (2) - 2

Decimal

x = - 0.73814 \dots

Solution steps

2 x - (lo g_{3} (4) + x + 1) + 3 = 0

Move

3

to the right side

2 x - (lo g_{3} (4) + x + 1) = - 3

Expand

2 x - (lo g_{3} (4) + x + 1) : x - lo g_{3} (4) - 1

x - lo g_{3} (4) - 1 = - 3

Move

lo g_{3} (4)

to the right side

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

Move

1

to the right side

x = 2 lo g_{3} (2) - 2

5 x + 4 = 2 x + 1

2 x + 7 = 17

x + 4 = - 9 - 3 (x - 5) + 8 x

a, \frac{a}{b + c} = \frac{a}{b} + \frac{a}{c}

y = - 2 (1) + 3

What is the answer to 2x-(log_{3}(4)+x+1)+3=0 ?
The answer to 2x-(log_{3}(4)+x+1)+3=0 is x=2log_{3}(2)-2
Solve for x: 2x-(log_{3}(4)+x+1)+3=0
The solution is x=2log_{3}(2)-2