What is log_{2}(16.0312+1)+log_{2}(16.0312-1) ?

The solution to log_{2}(16.0312+1)+log_{2}(16.0312-1) is log_{2}(399999021/1562500)

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simplify log_{2}(16.0312+1)+log_{2}(16.0312-1)

simplify

lo g_{2} (16.0312 + 1) + lo g_{2} (16.0312 - 1)

lo g_{2} (\frac{399999021}{1562500})

Decimal Notation

7.99999 \dots

Solution steps

lo g_{2} (16.0312 + 1) + lo g_{2} (16.0312 - 1)

Apply log rule:

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

lo g_{2} (16.0312 + 1) + lo g_{2} (16.0312 - 1) = lo g_{2} ((16.0312 + 1) (16.0312 - 1))

= lo g_{2} ((16.0312 + 1) (16.0312 - 1))

(16.0312 + 1) (16.0312 - 1) = 255.99937 \dots

= lo g_{2} (255.99937 \dots)

255.99937 \dots = \frac{25599937344}{100000000}

= lo g_{2} (\frac{25599937344}{100000000})

Cancel

\frac{25599937344}{100000000} : \frac{399999021}{1562500}

= lo g_{2} (\frac{399999021}{1562500})

ln (\frac{a ^{3}}{2} x^{2} e^{- 3 a})

ln (- 0.333) - ln (- \frac{1.2502}{3})

ln ((1 + b t) e^{- b t})

ln (3 e^{3}) + ln (\frac{2}{3 e ^{2}})

ln (\frac{\frac{1}{2}}{1})

What is log_{2}(16.0312+1)+log_{2}(16.0312-1) ?
The solution to log_{2}(16.0312+1)+log_{2}(16.0312-1) is log_{2}(399999021/1562500)