How do you solve #7-2e^x=5#?

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Answer 1 · 2015-05-22T23:52:15

Using #ln# is a useful way!

First, isolate #e^x# in one side:

#e^x=(5-7)/(-2)#
#e^x=1#

Now, apply #ln# on both sides, to remove #x# from the exponent, following the logarithm property #lna^n=n*lna#

#lne^x=ln1# (use a calculator for #ln=1#)
#xlne=ln1#

Now we have two definitions of #ln#:

#lne=1# (because #e^1=e#)
and #ln1=0# (because #e^0=1#)

Thus,

#x*1=0#
#x=0#