What is the answer to 128=16*log_{2}(1+x) ?

The answer to 128=16*log_{2}(1+x) is x=255

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128=16*log_{2}(1+x)

128 = 16 \cdot lo g_{2} (1 + x)

x = 255

Solution steps

128 = 16 lo g_{2} (1 + x)

Switch sides

16 lo g_{2} (1 + x) = 128

Divide both sides by

16

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

Apply log rules

1 + x = 256

Solve

1 + x = 256 : x = 255

x = 255

Verify Solutions:

x = 255

True

The solution is

x = 255

lo g_{x} (2) = 25

lo g_{x} (2) = 12

ln (t) = 1.56 + 0.82 \cdot ln (11)

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

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

What is the answer to 128=16*log_{2}(1+x) ?
The answer to 128=16*log_{2}(1+x) is x=255
Solve for x: 128=16*log_{2}(1+x)
The solution is x=255