How to solve the equation #log_(x) 5+log_(5) x=5/2# ?

Question

1277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-25T13:56:42

#x_1=sqrt5# and #x_2=25#

#log_x5+log_5x=5/2#

#log_x5+1/log_x5=5/2#

After using #y=log_x5#, this equation became

#y+1/y=5/2#

#(y^2+1)/y=5/2#

#2*(y^2+1)=5*y#

#2y^2+2=5y#

#2y^2-5y+2=0#

#(2y-1)*(y-2)=0#

From this equation, #y_1=1/2# and #y_2=2#

For #y=2#, #log_x5=2# or #x^2=5#. So #x_1=sqrt5#

For #y=1/2#, #log_x5=1/2# or #x^(1/2)=5#. So #x_2=5^2=25#