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

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Answer 1 · 2017-06-24T23:00:44

x = 81

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!