Question 206898
If this is whagt you meant:
{{{log(6,sqrt(x)) + log(6,sqrt (6x+5)) = 1}}}
.
Then, applying "log rules" we get:
{{{log(6,sqrt(x)sqrt(6x+5)) = 1}}}
{{{log(6,sqrt(6x^2+5x)) = 1}}}
{{{sqrt(6x^2+5x) = 6^1}}}
{{{sqrt(6x^2+5x) = 6}}}
{{{6x^2+5x = 36}}}
{{{6x^2+5x-36 = 0}}}
{{{6x^2+5x-36 = 0}}}
Solving the above with the quadratic equation yields the following:
x = {2.068, -2.901}
We can toss out the negative solution leaving:
x = 2.068
.
Details of quadratic follows:
*[invoke quadratic "x", 6, 5, -36 ]