How do you solve #log(x) + log(2x) = 10 #?

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Answer 1 · 2018-05-17T15:41:14

#color(maroon)(x = 50000sqrt2, -50000sqrt2#

#log m + log n = log (mn)#

Given : #log x + log (2x) = 10#

#:. log (x*2x) = 10#

#log (2x^2) = 10#

#2x^2 = 10^(10)#

#x^2 = 10^(10)/2#

#x = sqrt (10^(10)/2)#

#x = +-10^5/sqrt2 #

#color(maroon)(x = 50000sqrt2, -50000sqrt2#