How do you solve #e^x<4.5#?

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Answer 1 · 2018-04-16T11:08:16

#color(blue)((-oo, ln(9/2))#

#e^x<9/2#

We can express #9/2# as #e^(ln(9/2))#

#:.#

#e^x< e^(ln(9/2))#

For:

#a^x< a^y#

#x < y# if #a >1#

#:.#

#e > 1#

So:

#x< ln(9/2)#

#color(blue)((-oo, ln(9/2))#