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

The answer to 20=3log_{2}(1+s/n) is s=64*2^{2/3}n-n {n 0}

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20=3log_{2}(1+s/n)

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

s = 64 \cdot 2^{\frac{2}{3}} n - n {n < 0 or n > 0}

Solution steps

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

Switch sides

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

Divide both sides by

3

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

Apply log rules

1 + \frac{s}{n} = 2^{\frac{20}{3}}

Multiply both sides by

n

1 \cdot n + \frac{s}{n} n = 2^{\frac{20}{3}} n

Simplify

2^{\frac{20}{3}} n : 2^{6 + \frac{2}{3}} n

n + s = 2^{6 + \frac{2}{3}} n

Solve

n + s = 2^{6 + \frac{2}{3}} n : s = 64 \cdot 2^{\frac{2}{3}} n - n

s = 64 \cdot 2^{\frac{2}{3}} n - n

Verify Solutions:

s = 64 \cdot 2^{\frac{2}{3}} n - n {n < 0 or n > 0}

The solution is

s = 64 \cdot 2^{\frac{2}{3}} n - n {n < 0 or n > 0}

ln (\frac{t - 65}{125}) = - 0.07 t

lo g_{a} (x) = lo g_{a} (x)

y, ln (y) = 4 ln (x) + c

lo g_{10} ((x + 2) (x - 1)) = 1

1.73 = - lo g_{10} (x)

What is the answer to 20=3log_{2}(1+s/n) ?
The answer to 20=3log_{2}(1+s/n) is s=64*2^{2/3}n-n {n<0\lor n>0}