Question

Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)= In(3x/0); division by 0, undefined

Answer 1

Since 0 in ln(3x) - 0 is not a logarithm, the property of logarithms cannot be used here.

The difference shown cannot be written as a quotient of logarithms.

The step ln(x2) = ln(3x) - (0) reduces to

ln(x2) = ln(3x).

The possible solutions are 0 and 3, with 0 being extraneous.

right on edge

Answer 2

Error:
`lnx^2=ln 3x` not
`ln(3x)/(0)`

Solution:x=0 and 3

Explanation:

We have to find the error and correct answer

Given:
`2ln x=ln(3x)-[ln9-2ln(3)]`

`lnx^2=ln(3x)-[ln9-ln3^2]`

Using the formula

`alog b=logb^a`

`lnx^2=ln(3x)-[ln9-ln9]`

`lnx^2=ln(3x)-0`

`lnx^2=ln(3x)`

`x^2=3x`

`x^2-3x=0`

`x(x-3)=0`

Therefore, x=0 and x=3

But last step in the given solution

`lnx^2=ln(3x)/(0)=\infty`

It is wrong this property is used when

`log m-log n` then

`log(m)/(n)`

Hence, the student wrote
`lnx^2=ln(3x)/(0)`instead of
`lnx^2=ln3x` and solution is given by

x=0 and x=3