How do you solve #−10 + log_ 3 (n + 3) = −10#?

1 Answer
Lotusbluete
Jan 25, 2016

#n = -2#

Explanation:

#-10 + log_3(n+3) = -10#

First of all, add #10# on both sides of the equation:

#<=> color(white)(xx) log_3(n+3) = 0#

To "get rid" of the #log_3# term, you need to exponentiate the expression to the base #3#, since #a^x# is the inverse function for #log_a(x)# and thus, both #a^(log_a(x)) = x# and #log_a(a^x) = x# hold.

#<=> color(white)(xx) 3^(log_3(n+3)) = 3^0#

... don't forget that for any number #b# you can compute #b^0 = 1#

#<=> color(white)(xxxxx) n + 3 = 1#

... subtract #3# on both sides of the equation...

#<=> color(white)(xxxxxx) n = -2#

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