How do you solve $- 5 \cdot 2^{- 7 n - 7} + 5 = - 66$ $-5*2^(-7n-7)+5=-66$ ?

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Answer 1 · 2017-03-04T05:43:15

$n = - 1 - \sqrt[7]{{log}_{2} (\frac{71}{5})}$ $n=-1-root(7) (log_2 (71/5))$

You would subtract 5 and divide all by -5:

$- 5 \cdot 2^{- 7 n - 7} = - 71$ $-5*2^(-7n-7)=-71$

$2^{- 7 n - 7} = \frac{71}{5}$ $2^(-7n-7)=71/5$

then it is

$- 7 n - 7 = {log}_{2} (\frac{71}{5})$ $-7n-7=log_2 (71/5)$

$- 7 n = 7 + {log}_{2} (\frac{71}{5})$ $-7n=7+log_2 (71/5)$

$n = - 1 - \frac{1}{7} {log}_{2} (\frac{71}{5})$ $n=-1-1/7log_2 (71/5)$

$n = - 1 - \sqrt[7]{{log}_{2} (\frac{71}{5})}$ $n=-1-root(7) (log_2 (71/5))$

How do you solve -5*2^(-7n-7)+5=-66−5⋅2−7n−7+5=−66-5*2^(-7n-7)+5=-66?