How do you solve #\ln e ^ { x } = 3#?

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Answer 1 · 2018-04-19T00:45:46

#x=3#

Remember that:

#log_a(b^c)=clog_a(b)#

Therefore:

#=>log_e(e^x)=3#

#=>xlog_e(e)=3#

#=>x*1=3#

#=>x=3#

Answer 2 · 2018-04-19T00:48:32

#x=3#

A way to very quickly see the answer without much math at all is to convert the logarithmic expression into an exponential one, that is:

#e^x=e^3#

...based off of what a logarithm means, and that the natural logarithm is just a logarithm with base #e#.

A more conventional approach could be to use the end-around-multiplier approach (as you may have heard it), where:

#ln(e^x)=3#
#x*ln(e)=3#
#x*1=x=3#

Answer 3 · 2018-04-20T13:35:58

#x=3#

Given: #lne^x=3#.

#=>xlne=3# #(because lna^b=blna)#

#=>x*1=3# #(because lne=1)#

#=>x=3#