What is 1+2log_{2}(pi)-log_{2}(15) ?

The solution to 1+2log_{2}(pi)-log_{2}(15) is 1+log_{2}((pi^2)/(15))

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simplify 1+2log_{2}(pi)-log_{2}(15)

simplify

1 + 2 lo g_{2} (π) - lo g_{2} (15)

1 + lo g_{2} (\frac{π ^{2}}{15})

Decimal Notation

0.39610 \dots

Solution steps

1 + 2 lo g_{2} (π) - lo g_{2} (15)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{2} (π) = lo g_{2} (π^{2})

= 1 + lo g_{2} (π^{2}) - lo g_{2} (15)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{2} (π^{2}) - lo g_{2} (15) = lo g_{2} (\frac{π ^{2}}{15})

= 1 + lo g_{2} (\frac{π ^{2}}{15})

4186 (ln (323) - ln (373))

ln (\frac{51.65}{50.35})

ln (50) \cdot ln (\frac{1}{50})

ln (x y^{8} z^{9})

ln (ln (x)) - ln (f (x))

What is 1+2log_{2}(pi)-log_{2}(15) ?
The solution to 1+2log_{2}(pi)-log_{2}(15) is 1+log_{2}((pi^2)/(15))