What is log_{2}(7/11 e^{15 1/4}) ?

The solution to log_{2}(7/11 e^{15 1/4}) is log_{2}(7)-log_{2}(11)+15 1/4 log_{2}(e)

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simplify log_{2}(7/11 e^{15 1/4})

simplify

lo g_{2} (\frac{7}{11} e^{15 \frac{1}{4}})

lo g_{2} (7) - lo g_{2} (11) + 15 \frac{1}{4} lo g_{2} (e)

Decimal Notation

21.34902 \dots

Solution steps

lo g_{2} (\frac{7}{11} e^{15 \frac{1}{4}})

Convert mixed numbers to improper fractions:

15 \frac{1}{4} = \frac{61}{4}

= lo g_{2} (\frac{7}{11} e^{\frac{61}{4}})

Apply log rule:

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

lo g_{2} (\frac{7}{11} e^{\frac{61}{4}}) = lo g_{2} (\frac{7}{11}) + lo g_{2} (e^{\frac{61}{4}})

= lo g_{2} (\frac{7}{11}) + lo g_{2} (e^{\frac{61}{4}})

Apply log rule:

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

lo g_{2} (e^{\frac{61}{4}}) = \frac{61}{4} lo g_{2} (e)

= lo g_{2} (\frac{7}{11}) + \frac{61}{4} lo g_{2} (e)

Apply log rule:

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

lo g_{2} (\frac{7}{11}) = lo g_{2} (7) - lo g_{2} (11)

= lo g_{2} (7) - lo g_{2} (11) + \frac{61}{4} lo g_{2} (e)

Convert improper fractions to mixed numbers:

\frac{61}{4} = 15 \frac{1}{4}

= lo g_{2} (7) - lo g_{2} (11) + 15 \frac{1}{4} lo g_{2} (e)

ln (\frac{273}{259.65})

2770 ln (\frac{59}{1 0 ^{11} \cdot 1550})

3 ln (x) + 3 + ln (e x) + 1

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

ln (1 + 2) \cdot \frac{1}{2} - ln (- 12)

What is log_{2}(7/11 e^{15 1/4}) ?
The solution to log_{2}(7/11 e^{15 1/4}) is log_{2}(7)-log_{2}(11)+15 1/4 log_{2}(e)