Question #8acfd

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Question #8acfd

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Answer 1 · 2015-06-18T08:52:03

The value of #n# is #root(5)(10)/5#

Explanation:

When you substitute values for x,a and b you get:
#2=10log_10 ((25n)/5)#
#2=10log_10 5n#
#1/5=log_10 5n#
#log_10 (10^(1/5))=log_10 5n#
#10^(1/5)=5n#
#n=10^(1/5)/5#

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Question #8acfd

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