How do you solve #\log ( x + 19) = 2#?

1 Answer
Jun 24, 2017

x = 81

Explanation:

When no base is written, it's assumed that the base is 10 .

#log(x + 19) = 2#
#log_10(x + 19) = 2#

A general rule for logs is that if #log_b(a) = c#, then #b^c = a#.

Therefore, we know that

#log_10(x + 19) = 2#
#10^2 = x + 19#

To find x, simplify the equation.

#100 = x + 19#
#81 = x#

Hope this helps!