How to solve #log _{x}36+log _{18}3x = 3# ?

Question

#log _{x}36+log _{18}3x = 3#

Benjamin Achtenberg · Accepted Answer

#x in {324; 6}# #frac{log_2 36}{log_2 x} + frac{log_2 3 + log_2 x}{log_2 18} = 3# #log_2 x = y# #log_2 3 = K# #frac{log_2 18 log_2 36}{y} + K + y = 3 log_2 18# #frac{log_2 18 log_2 36}{y} + y = 3 (1 + 2 K) - K# #c + y^2 = - by # #b = - 3 - 5K# #c = (1 + 2 K)(2 + 2 K) = 4K^2 + 6K + 2# #Delta = (5K + 3)^2 - 16K^2 - 24K - 8# #Delta = 9K^2 + 6K + 1# #y = log_2 x = (3 + 5K pm (3K + 1))/2# #x_1 =2 wedge (2 + 4K) = 4 * 3^4# #x_2 =2 wedge (1 + K) = 2 * 3#

James Allen · Answer

#color(blue)(x=324, x=6)#

#log_x36=log_18(36)/(log_18(x)# (change of base)

#log_18(36)/(log_18(x))+log_18(3x)=3#

#(2log_18(6))/(log_18(x))+log_18(x)+log_18(3)=3#

Let #m=log_18(x)#

#(2log_18(6))/m+m+log_18(3)=3#

#(2log_18(6))+m^2+log_18(3)m=3m#

#m^2+m(log_18(3)-3)+2log_18(6)=0#

Using quadratic formula:

#m=(-(log_18(3)-3)+-sqrt(((log_18(3)-3))^2-4(2log_18(6))))/2#

#m=(-(log_18(3)-3)+ sqrt(((log_18(3)-3))^2-(8log_18(6))))/2#

#=2#

#m=(-(log_18(3)-3)-sqrt(((log_18(3)-3))^2-(8log_18(6))))/2#

#=0.6199062340#

#m=log_18(x)#

#log_18(x)=2#

#18^(log_18(x))=18^2#

#x=18^2=324#

#log_18(x)=0.6199062340#

#18^(log_18(x))=18^(0.6199062340)#

#x=18^(0.6199062340)=6#

James Allen · Answer

undefined To solve the equation $ \log_x 36 + \log_{18} 3x = 3 $, you can use the properties of logarithms.