What is the answer to 20=3log_{2}(1+x) ?

The answer to 20=3log_{2}(1+x) is x=64*2^{2/3}-1

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20=3log_{2}(1+x)

20 = 3 lo g_{2} (1 + x)

x = 64 \cdot 2^{\frac{2}{3}} - 1

Decimal Notation

x = 100.59366 \dots

Solution steps

20 = 3 lo g_{2} (1 + x)

Switch sides

3 lo g_{2} (1 + x) = 20

Divide both sides by

3

lo g_{2} (1 + x) = \frac{20}{3}

Apply log rules

1 + x = 64 \cdot 2^{\frac{2}{3}}

Solve

1 + x = 64 \cdot 2^{\frac{2}{3}} : x = 64 \cdot 2^{\frac{2}{3}} - 1

x = 64 \cdot 2^{\frac{2}{3}} - 1

Verify Solutions:

x = 64 \cdot 2^{\frac{2}{3}} - 1

True

The solution is

x = 64 \cdot 2^{\frac{2}{3}} - 1

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

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

lo g_{2} (x) - 3 lo g_{2} (x) = 6

lo g_{x} (2) = 0

lo g_{x} (2) = 5

What is the answer to 20=3log_{2}(1+x) ?
The answer to 20=3log_{2}(1+x) is x=64*2^{2/3}-1
Solve for x: 20=3log_{2}(1+x)
The solution is x=64*2^{2/3}-1