What is log_{2}(e^9)-log_{2}(e^6) ?

The solution to log_{2}(e^9)-log_{2}(e^6) is 3log_{2}(e)

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simplify log_{2}(e^9)-log_{2}(e^6)

simplify

lo g_{2} (e^{9}) - lo g_{2} (e^{6})

3 lo g_{2} (e)

Decimal Notation

4.32808 \dots

Solution steps

lo g_{2} (e^{9}) - lo g_{2} (e^{6})

Apply log rule:

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

lo g_{2} (e^{9}) - lo g_{2} (e^{6}) = lo g_{2} (\frac{e ^{9}}{e ^{6}})

= lo g_{2} (\frac{e ^{9}}{e ^{6}})

\frac{e ^{9}}{e ^{6}} = e^{3}

= lo g_{2} (e^{3})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= 3 lo g_{2} (e)

lo g_{5} (8 \cdot 11 \cdot 11 \cdot 3^{3})

5 (- ln ∣ 1 - 4 ∣ + ln ∣ 1 - 5 ∣) + y + C

ln (\frac{2 ^{0.538 - 1} - 1}{( 0.538 - 1 ) x})

ln (e^{t} \cdot ln (t))

\frac{1}{2} (ln (11) - ln (3))

What is log_{2}(e^9)-log_{2}(e^6) ?
The solution to log_{2}(e^9)-log_{2}(e^6) is 3log_{2}(e)