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RoseWind [281]
3 years ago
9

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
Mathematics
1 answer:
Art [367]3 years ago
4 0

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

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