Question

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

Answer 1

`x = +-10`

Explanation:

`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.