What is e^{(ln(3)-ln(8)+ln(25)-ln(6))/2} ?

The solution to e^{(ln(3)-ln(8)+ln(25)-ln(6))/2} is e^{(ln(25/16))/2}

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simplify e^{(ln(3)-ln(8)+ln(25)-ln(6))/2}

simplify

e^{\frac{l n ( 3 ) - l n ( 8 ) + l n ( 25 ) - l n ( 6 )}{2}}

e^{\frac{l n ( \frac{25}{16} )}{2}}

Decimal Notation

1.25

Solution steps

e^{\frac{l n ( 3 ) - l n ( 8 ) + l n ( 25 ) - l n ( 6 )}{2}}

Apply log rule:

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

ln (3) - ln (8) = ln (\frac{3}{8})

= e^{\frac{l n ( \frac{3}{8} ) + l n ( 25 ) - l n ( 6 )}{2}}

Apply log rule:

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

ln (\frac{3}{8}) + ln (25) = ln (25 \cdot \frac{3}{8})

= e^{\frac{l n ( \frac{3}{8} \cdot 25 ) - l n ( 6 )}{2}}

\frac{3}{8} \cdot 25 = \frac{75}{8}

= e^{\frac{l n ( \frac{75}{8} ) - l n ( 6 )}{2}}

Apply log rule:

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

ln (\frac{75}{8}) - ln (6) = ln (\frac{\frac{75}{8}}{6})

= e^{\frac{l n ( \frac{\frac{75}{8}}{6} )}{2}}

\frac{\frac{75}{8}}{6} = \frac{25}{16}

= e^{\frac{l n ( \frac{25}{16} )}{2}}

lo g_{10} (\frac{5 x + 7}{2})

lo g_{2} (8 d)

lo g_{h} (9 jk)

lo g_{10} (\frac{2}{λ} (1 - y)^{\frac{( 2 - λ )}{λ}})

What is e^{(ln(3)-ln(8)+ln(25)-ln(6))/2} ?
The solution to e^{(ln(3)-ln(8)+ln(25)-ln(6))/2} is e^{(ln(25/16))/2}