Question 1114883
<br>
{{{log(5x+9,(x^2+6x+9))+log(x+3,(5x^2+24x+27))=4}}}<br>
{{{log(5x+9,((x+3)^2))+log(x+3,((x+3)(5x+9)))=4}}}<br>
{{{log(5x+9,((x+3)^2))+log((x+3),(x+3))+log((x+3),(5x+9))=4}}}<br>
{{{log(5x+9,((x+3)^2))+1+log((x+3),(5x+9))=4}}}<br>
{{{log(5x+9,((x+3)^2))+log((x+3),(5x+9))=3}}}<br>
{{{2*log(5x+9,(x+3))+log((x+3),(5x+9))=3}}}<br>
Make a u substitution to make the notation easier....<br>
{{{u = log(5x+9,x+3)}}}<br>
Then<br>
{{{1/u = log(x+3,5x+9)}}}<br>
Then the equation is<br>
{{{2u + 1/u = 3}}}
{{{2u^2+1 = 3u}}}
{{{2u^2-3u+1 = 0}}}
{{{(2u-1)(u-1) = 0}}}
{{{u = 1/2}}} or {{{u = 1}}}<br> 
(1) {{{log(5x+9,x+3) = 1/2}}}
or
(2) {{{log(5x+9,x+3)=1}}}<br>
See where each of those solutions leads us....<br>
(1) {{{(5x+9)^(1/2) = x+3}}}
{{{5x+9 = x^2+6x+9}}}
{{{x^2+x = 0}}}
{{{x(x+1) = 0}}}
{{{x = 0}}} or {{{x = -1}}}<br>
Check x=0: x+3 = 3; 5x+9 = 9
{{{log(9,9)+log(3,(3*9)) = 1+3 = 4}}} CHECK!<br>
Check x=-1: x+3 = 2; 5x+9 = 4
{{{log(4,4)+log(2(2*4)) = 1+3 = 4}}} CHECK!<br>
(2) {{{(5x+9)^1 = x+3}}}
{{{5x+9 = x+3}}}
{{{4x = -6}}}
{{{x = -3/2}}}<br>
Check x = -3/2: x+3 = 3/2; 5x+9 = 3/2
{{{log(3/2,(3/2)^2)+log(3/2,((3/2)*(3/2))) = 2+2 = 4}}} CHECK!<br>
Answer: There are 3 solutions: x=0, x=-1, and x=-3/2.