Asked by Jillian
Simplify completely.
e^(ln(7)+2ln(4))−log9(27/27)
the log has a base of 9 and the top 27 is under a radical sign.. my computer doesn't have symbols for them.
the base of e is raised to In(7)+2In(4)
e^(ln(7)+2ln(4))−log9(27/27)
the log has a base of 9 and the top 27 is under a radical sign.. my computer doesn't have symbols for them.
the base of e is raised to In(7)+2In(4)
Answers
Answered by
Steve
e^(ln7 + 2ln4)
= e^(ln7+ln16)
= e^ln112
= 112
√27/27 = 1/√27 = 27^(-1/2) = (9^3)^(-1/2) = 9^(-3/2)
log_9(9^-3/2) = -3/2
remember that
log_b(b^n) = n
b^log_b(n) = n
that is the definition of the log.
= e^(ln7+ln16)
= e^ln112
= 112
√27/27 = 1/√27 = 27^(-1/2) = (9^3)^(-1/2) = 9^(-3/2)
log_9(9^-3/2) = -3/2
remember that
log_b(b^n) = n
b^log_b(n) = n
that is the definition of the log.
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