What is the answer to 2log_{2}(4x+1)-1=6 ?

The answer to 2log_{2}(4x+1)-1=6 is x=(8sqrt(2)-1)/4

Solve for x: 2log_{2}(4x+1)-1=6

The solution is x=(8sqrt(2)-1)/4

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2 lo g_{2} (4 x + 1) - 1 = 6

x = \frac{8 2 - 1}{4}

Decimal Notation

x = 2.57842 \dots

Solution steps

2 lo g_{2} (4 x + 1) - 1 = 6

Move

1

to the right side

2 lo g_{2} (4 x + 1) = 7

Divide both sides by

2

lo g_{2} (4 x + 1) = \frac{7}{2}

Apply log rules

4 x + 1 = 82

Solve

4 x + 1 = 82 : x = \frac{8 2 - 1}{4}

x = \frac{8 2 - 1}{4}

Verify Solutions:

x = \frac{8 2 - 1}{4}

True

The solution is

x = \frac{8 2 - 1}{4}

lo g_{6} (x + 27) - lo g_{6} (x - 8) = 2

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

lo g_{3} (9 x) = 5 - lo g_{3} (8)

y, lo g_{e} (y) = x

lo g_{2} (- 4 x^{2} + 12 x) = 3

What is the answer to 2log_{2}(4x+1)-1=6 ?
The answer to 2log_{2}(4x+1)-1=6 is x=(8sqrt(2)-1)/4
Solve for x: 2log_{2}(4x+1)-1=6
The solution is x=(8sqrt(2)-1)/4