How do you solve #4 log x - log x = 64#?

Question

2204 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-21T20:29:09

#x=10^(64/3)#

#4logx-logx=64#

#3logx=64#

#logx=64/3#

Recall that #logx=log_10x#.

Exponentiate both sides to undo the logarithm.

#10^(log_10x)=10^(64/3)#

#x=10^(64/3)#

#x~~2.1544xx10^21#