Question

How do you solve `log_3 x = 9 log_x 3`?

Answer 1

Using the change of base rule you get

#logx/log3=9log3/logx=>(logx)^2=(3log3)^2=> (logx-3log3)*(logx+3log3)=0=> x=3^3=27 or x=1/27#