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Question #d5146

1 Answer

Aug 29, 2017

#x=0.633# to 3 s.f.

Explanation:

Let's put logs in and rearrange equation to have #x# on its own.

#5xlog7 = log475 -> 5x=log475/log7#

#5x = 3.1673#

Divide by 5 to get #x# on its own:

#x=3.1673/5#

#x=0.633# to 3 s.f.

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