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

Answer:

Error:$lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step 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

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote  $lnx^2=ln\frac{3x}{0}$instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

Answer 2

Answer:

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