How do you evaluate #\ln e ^ { 8} - 7\ln e ^ { 3} #?

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Answer 1 · 2018-04-11T12:05:14

#13#

logarithm rule:

#log a^n = n log a#

this also applies for #log_e: ln a^n = n ln a#.

using this rule, #ln e^8# is the same as #8 ln e#.

#ln e# is the power that #e# is raised by to equal #e#. this power is #1#.

#ln e# is #1#, so #8 ln e# is #8#.

#ln e^3# is the same as #3 ln e#.

#7 ln e^3# is the same as #7 * 3 ln e#.

#7 * 3 = 21#, so #7 ln e^3# is #21 ln e#.

#(8 ln e) - (21 ln e)# is #8 - 21#.

#8 - 21 = -13#

hence, #ln e^8 - 7 ln e^3 = 8 - 21 = -13#.