How do you solve #log 5 + log x = 2#?

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Answer 1 · 2017-09-29T08:00:05

#x=20#

assuming #logx# is base #10#

we have

#log5+logx=2#

using the property

#logA+logB=logAB#

log5x=2#

using the definition

#log_ab=c=>a^c=b#

#10^2=5x#

#+>5x=100#

#:.x=20#