How do you solve #log_3x-2log_x3=1#?

3 Answers
Mar 6, 2018

See the subscript

Explanation:

Handwritten

Mar 6, 2018

#x = 1/3, 9#

Explanation:

#log_b a = 1/log_a b#

#log_3x−2log_x3=1#

#log_3x−2/log_3x=1#

Let log_3x = a

#a - 2/a = 1#

#a^2 - a - 2 = 0#

#(a - 2)(a + 1) = 0#

#a = -1, a = 2#

=> #log_3x = -1 or log_3x = 2#

#log_3x = -1 => x = 3^-1 = 1/3#
or
# log_3x = 2 => x = 3^2 = 9#

#x = 1/3, 9#

Mar 6, 2018

#x=9 and x=1/3#

Explanation:

Note that #2log_x(3)->log_x(3^2)=log_x(9)#

#log_3(x)-log_x(9)=1#

#log_3(x)=1+log_x(9)#

Consider the LHS #->log_10(x)/log_10(3) " "...Equation(1) =y_1#

Consider the RHS #->1+log_10(9)/(log_10(x))" "..Equation(2)=y_2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As we have got all #log_10# now we can drop the base 10 and use just log.

Try #y_1=y_2# and see what happens.

#log(x)/log(3)=1+log(9)/(log(x))#

#log(x) /log(3) = (log(x)+log(9))/log(x)#

Cross multiply

#log(x)^2=log(x)log(3)+log(3)log(9)#

Set #log(x) = t# giving

#t^2-tlog(3)-log(3)log(9)=0 larr" A quadratic in "t#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#y=ax^2+bx+c -> x=(-b+-sqrt(b^2-4ac))/(2a)#

Where #a=1; b=-log(3); c= -log(3)log(9)#

#t=(+log(3)+-sqrt(log(3)^2-4(1)(-log(3)log(9))))/(2(1))#

#t=+log(3)/2+-sqrt(2.0488022...)/2#

#t~~+0.95424...=>log^-1(t)= 9 = x#

#t~~-0.4771212....=>log^(-1)(t)= 1/3=x#

Tony B