How do you solve #log_3[3^(x^2-13x+28)+2/9] = log_2 5#?

1 Answer
Jun 25, 2016

Answer:

#x =13/2 pm 1/2 sqrt[49 + (4 Log_e(3^(Log_e[20]/Log_e[2])-2))/Log_e 3]#

Explanation:

#log_3(3^(x^2-13x+28)+2/9)= log_2 5#
#3^{log_3(3^(x^2-13x+28)+2/9)}=3^{ log_2 5}#
#3^{x^2-13x+28}+2/9=3^{ log_2 5}#
#3^2 xx 3^{x^2-13x+28}+2=3^2 xx 3^{ log_2 5}#
#3^{x^2-13x+30}+2=3^{ log_2 5 + 2}#
#3^{x^2-13x+30}=3^{ log_2 5 + 2}-2#

Solving for #x#

#x =13/2 pm 1/2 sqrt[49 + (4 Log_e(3^(Log_e[20]/Log_e[2])-2))/Log_e 3]#