#log_9 (3u + 14) - log_3(2u) = log_5 5#
We now use the change of base formula #log_a n = logn/loga#.
#log(3u + 14)/log9 - log(2u)/log3 = log5/log5#
#log(3u + 14)/log9 - log(2u)/log3 = 1#
Now apply #loga^n = nloga#.
#log(3u + 14)/(2log3) - log(2u)/log3 = 1#
#log(3u + 14)/(2log3) - (2log(2u))/(2log3) = 1#
#log_9 (3u + 14) - 2log_9 (2u) = 1#
#log_9 (3u + 14) - log_9 (4u^2) = 1#
Now apply #log_a n - log_a m = log_a (n/m)#
#log_9 ((3u + 14)/(4u^2)) = 1#
#(3u + 14)/(4u^2) = 9#
#3u + 14 = 36u^2#
#0 = 36u^2 - 3u - 14#
#u = (-(-3) +- sqrt((-3)^2 - 4 * 36 *- 14))/(2*72)#
#u = (3 +- sqrt(2025))/144#
#u = 48/144 or -42/144#
#u = 1/3 or -7/24#
However, #log_2(na)# can never take a negative value of #a# as long as #n# is positive, therefore the only solution is #u = 1/3#.
Hopefully this helps!
