How do you solve $- 1 - 8 ln (- 8 b - 5) = 7$ $-1-8ln(-8b-5)=7$ ?

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Answer 1 · 2018-04-30T17:41:01

$b = \frac{1}{8} (\frac{1}{e} + 5)$ $b = 1/8(1/e + 5)$

$- 1 - 8 ln (- 8 b - 5) = 7$ $-1-8ln(-8b-5)=7$

$ln (- 8 b - 5) = - 1$ $ln(-8b-5)= -1$

$Note : {log}_{b} a = c \Rightarrow a = b^{c}$ $"Note" : log_b a = c => a = b^c$

$- 8 b - 5 = e^{- 1}$ $-8b-5 = e^-1$

$b = \frac{1}{8} (\frac{1}{e} + 5)$ $b = 1/8(1/e + 5)$

How do you solve -1-8ln(-8b-5)=7−1−8ln(−8b−5)=7-1-8ln(-8b-5)=7?