Question

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

Answer 1

See the subscript

Explanation:

Handwritten

Answer 2

`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`

Answer 3

`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