How do you solve the equation #log_3 0.1+2log_3x=log_3 2+log_3 5#?

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Answer 1 · 2016-10-19T17:45:46

#x = +-10#

#log_3 0.1 + color(red)(2log_3 x) = log_3 2 + log_3 5" "larr# use the power law

#:.log_3 0.1 + color(red)(log_3 x^2) = log_3 2 + log_3 5#

#color(blue)(log_a b + log_a c = log_a (b xxc))" "larr# multiply law

#:.log_3 0.1 xx x^2 = log_3 2 xx 5#

#log_3 0.1x^2 = log_3 10#

#color(darkviolet)(log A = log B hArr A = B)#

#:. 0.1x^2 = 10#

#x^2 = 10/0.1#

#x^2 = 100#

#x = +-sqrt100#

#x = +10 " or " x = -10#

Usually log of a negative number is invalid, but in this case, even if #x = -10#, when it is squared it becomes positive.