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

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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 .

$log (x + 19) = 2$ $log(x + 19) = 2$
${log}_{10} (x + 19) = 2$ $log_10(x + 19) = 2$

A general rule for logs is that if ${log}_{b} (a) = c$ $log_b(a) = c$ , then $b^{c} = a$ $b^c = a$ .

Therefore, we know that

${log}_{10} (x + 19) = 2$ $log_10(x + 19) = 2$
$10^{2} = x + 19$ $10^2 = x + 19$

To find x, simplify the equation.

$100 = x + 19$ $100 = x + 19$
$81 = x$ $81 = x$

Hope this helps!

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