Question 1094820
<br>No solution.<br>
{{{log_4(x+6)+9 = 11-log_4(x+21)}}}
{{{log_4(x+6)+log_4(x+21) = 11-9 = 2}}}
{{{log_4(x+6)+log_4(x+21) = log_4(16)}}}
{{{(x+6)(x+21) = 16}}}
{{{x^2+27x+126 = 16}}}
{{{x^2+27x+110 = 0}}}
{{{(x+5)(x+22) = 0}}}<br>
The two solutions to the quadratic equation are x = -5 and x = -22.<br>
But both of those solutions, tried in the original equations, result in trying to evaluate the log of a negative number.