How do you solve #-3e^(9x-1)+6=-58#?

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Answer 1 · 2016-11-20T15:34:11

#x=1/9(ln(64/3)+1)#

#-3e^(9x-1)+6=-58#

Subtract 6 from each side:
#-3e^(9x-1)=-64#

Divide each side by -3:
#e^(9x-1)=64/3#

To isolate x, rewrite the equation as a natural log:
#9x-1=ln(64/3)#

Add 1 to each side:
#9x=ln(64/3)+1#

Divide each side by 9:
#x=1/9(ln(64/3)+1)#

This can be expanded and written as:
#x=1/9(ln64-ln3+1)#