How do you solve #7-4logx=10#?

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Answer 1 · 2015-12-19T14:00:09

#10^(-3/4)#

#7-4logx=10#

Subtract #7# from both sides.

#-4logx=3#

Divide both sides by #-4#.

#logx=-3/4#

Recall that #logx=log_10 x#.

Exponentiate both sides to undo the logarithm

#10^(log_10x)=10^(-3/4)#

#x=10^(-3/4)#

This can be written in a variety of different ways.

#10^(-3/4)=1/(10^(3/4))=1/root4(10^3)=1/root4(1000)~=0.1778#