Question 132012
{{{log(X) + log(X) + 1 = log(12)}}}
First get everything in terms of logs
1 = log(x)
If the base is {{{10}}}, then I can write
{{{10^1 = x}}}
{{{x = 10}}}, so
{{{1 = log(10)}}}
Now rewrite it
{{{log(X) + log(X) + log(10) = log(12)}}}
The general rule for logs is:
{{{log(a*b*c) = log(a) + log(b) + log(c)}}} so,
{{{log(X) + log(X) + log(10) = log(X*X*10)}}}
{{{log(10x^2) = log(12)}}}
If the logs are equal, the numbers are equal
{{{10X^2 = 12}}}
{{{X^2 = 6/5}}}
{{{X = sqrt(6/5)}}} answer 
There is a negative square root also, but it makes
no sense here, since there is no log to the base 10
that will give me a negative number
check:
{{{sqrt(6/5) = 1.0954}}}
{{{log(X) + log(X) + 1 = log(12)}}}
{{{log(1.0954) + log(1.0954) + 1 = log(12)}}}
{{{.03957 + .03957 + 1 = 1.0792}}}
{{{.07914 + 1 = 1.0792}}}
{{{1.07914 = 1.0792}}}
close enough