How do you solve #3e^(-3x) +4 = 6#?

1 Answer
Douglas K.
Mar 27, 2018

Given: #3e^(-3x) +4 = 6#

Subtract 4 from both sides:

#3e^(-3x) = 2#

Divide both sides by 3:

#e^(-3x) = 2/3#

Because the natural logarithm and the exponential function are inverses, using the natural logarithm on both sides will leave only the exponent on the left:

# -3x = ln(2/3)#

Multiply both sides by #-1/3#

# x = -1/3ln(2/3)#

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