What is 2log_{e}(9)+2log_{e}(pi)-log_{e}(8) ?

The solution to 2log_{e}(9)+2log_{e}(pi)-log_{e}(8) is ln((81pi^2)/8)

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simplify 2log_{e}(9)+2log_{e}(pi)-log_{e}(8)

simplify

2 lo g_{e} (9) + 2 lo g_{e} (π) - lo g_{e} (8)

ln (\frac{81 π ^{2}}{8})

Decimal Notation

4.60446 \dots

Solution steps

2 lo g_{e} (9) + 2 lo g_{e} (π) - lo g_{e} (8)

Apply log rule:

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

2 lo g_{e} (9) = lo g_{e} (9^{2})

= lo g_{e} (9^{2}) + 2 lo g_{e} (π) - lo g_{e} (8)

Apply log rule:

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

2 lo g_{e} (π) = lo g_{e} (π^{2})

= lo g_{e} (9^{2}) + lo g_{e} (π^{2}) - lo g_{e} (8)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{e} (9^{2}) + lo g_{e} (π^{2}) = lo g_{e} (9^{2} π^{2})

= lo g_{e} (9^{2} π^{2}) - lo g_{e} (8)

9^{2} = 81

= lo g_{e} (81 π^{2}) - lo g_{e} (8)

Apply log rule:

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

lo g_{e} (81 π^{2}) - lo g_{e} (8) = lo g_{e} (\frac{81 π ^{2}}{8})

= lo g_{e} (\frac{81 π ^{2}}{8})

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (\frac{81 π ^{2}}{8})

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

ln (\frac{47}{50}) + (0.085) \cdot 0.054

\frac{2 ln ( 5 )}{ln ( 2 )} - 3

- 2 ln (\frac{10 - x}{10})

(\frac{a}{b}) - 2 ln (a^{3}) + 4 ln (b)

What is 2log_{e}(9)+2log_{e}(pi)-log_{e}(8) ?
The solution to 2log_{e}(9)+2log_{e}(pi)-log_{e}(8) is ln((81pi^2)/8)