How do you solve #4e^(r+5)=29#?

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Answer 1 · 2016-10-02T02:15:34

#r=ln(29/4)-5color(white)(aaa)# or #color(white)(aa)r=-3.24214#

#4e^(r+5)=29#

#(4e^(r+5))/4=29/4color(white)(aaaa)# Divide both side by 4

#e^(r+5)=29/4#

#lne^(r+5)=ln(29/4)color(white)(aaaa)# Take the natural log of both sides

#(r+5)lne=ln(29/4)color(white)(aaa)# Use the log rule #logx^a=alogx#

#r+5=ln(29/4)color(white)(aaaaaaa)# Recall that #lne=1#

#r=ln(29/4)-5color(white)(aaaa)# Subtract 5 from both sides

#r=-3.24214color(white)(aaaa)# Use a calculator