What is the answer to d=log_{2}(8/(sqrt(d))) ?

The answer to d=log_{2}(8/(sqrt(d))) is d=(\W_{0}(128ln(2)))/(2ln(2))

Solve for d: d=log_{2}(8/(sqrt(d)))

The solution is d=(\W_{0}(128ln(2)))/(2ln(2))

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d = lo g_{2} (\frac{8}{d})

d = \frac{W _{0} ( 128 ln ( 2 ) )}{2 ln ( 2 )}

Decimal Notation

d = 2.37579 \dots

Solution steps

d = lo g_{2} (\frac{8}{d})

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{d} = 2^{l o g_{2} (\frac{8}{d})}

Simplify

2^{l o g_{2} (\frac{8}{d})} : \frac{8}{d}

2^{d} = \frac{8}{d}

Multiply both sides by

d

2^{d} d = \frac{8}{d} d

Simplify

2^{d} d = 8

Solve

2^{d} d = 8 : d = \frac{W _{0} ( 128 ln ( 2 ) )}{2 ln ( 2 )}

d = \frac{W _{0} ( 128 ln ( 2 ) )}{2 ln ( 2 )}

Verify Solutions:

d = \frac{W _{0} ( 128 ln ( 2 ) )}{2 ln ( 2 )}

True

The solution is

d = \frac{W _{0} ( 128 ln ( 2 ) )}{2 ln ( 2 )}

2 ln (x) - (ln (x))^{2} = 0

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

lo g_{3} (x - 4) + lo g_{3} (x + 6) = 1

lo g_{x} (1.41) = 0.5

- 3 = 20 lo g_{10} (x)

What is the answer to d=log_{2}(8/(sqrt(d))) ?
The answer to d=log_{2}(8/(sqrt(d))) is d=(\W_{0}(128ln(2)))/(2ln(2))
Solve for d: d=log_{2}(8/(sqrt(d)))
The solution is d=(\W_{0}(128ln(2)))/(2ln(2))