What is e^{(ln(1+1^4)-ln(17))/4} ?

The solution to e^{(ln(1+1^4)-ln(17))/4} is e^{(ln(2/17))/4}

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simplify e^{(ln(1+1^4)-ln(17))/4}

simplify

e^{\frac{l n ( 1 + 1 ^{4} ) - l n ( 17 )}{4}}

e^{\frac{l n ( \frac{2}{17} )}{4}}

Decimal

0.58565 \dots

Solution steps

e^{\frac{l n ( 1 + 1 ^{4} ) - l n ( 17 )}{4}}

Apply log rule:

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

ln (1 + 1^{4}) - ln (17) = ln (\frac{1 ^{4} + 1}{17})

= e^{\frac{l n ( \frac{1 + 1 ^{4}}{17} )}{4}}

\frac{1 + 1 ^{4}}{17} = \frac{2}{17}

= e^{\frac{l n ( \frac{2}{17} )}{4}}

2 lo g_{0.5} (- 80 (x - 90)) + 10

10 lo g_{10} (\frac{2 \cdot 1 0 ^{- 7}}{1 0 ^{- 12}})

ln (5^{x}) - ln (5^{x - 1})

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

\frac{ln ( \frac{29}{2} )}{5 ln ( 2 )}

What is e^{(ln(1+1^4)-ln(17))/4} ?
The solution to e^{(ln(1+1^4)-ln(17))/4} is e^{(ln(2/17))/4}