How do you solve #e^ { 1- 8x } = 7957#?

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Answer 1 · 2016-11-16T01:36:48

Using the inverse function ln we convert it to a simple algebraic expression.

Using the inverse function ln we convert it to a simple algebraic expression.
#ln (e^(1-8x) = ln (7957)# (look up ln(7957) or use a calculator)

1-8x = 8.98 ; -8x = 7.98 ; x = -0.9975

CHECK:
#e^(1-8*-0.9975) = (7957)#
#e^(1+7.98) = (7957)#
#e^(8.98) = (7957)# ; 7942 = 7957 (within error of approximation for 2 significant digits)

Answer 2 · 2016-11-16T01:43:29

#e^(1-8x)=7957# : Given

#ln e^(1-8x)= ln 7957# : Take the natural log of both sides to try to get a step closer to isolating x

#(1-8x) ln e= ln 7957# : ln e are inverses of each other, which equals 1. Bring the exponent #(1-8x)# to the front.

#1-8x=ln7957#

#-8x=ln 7957-1# : Subtract 1 to the other side to isolate x

#x=-(ln 7957 - 1)/8# : Divide by 8 to the other side to isolate x