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The approximate value of the expression 2log⁡33n−log⁡3(n2+1)2\log_3 3n - \log_3(n^2 + 1) for a sufficiently large nn is

Solution

✅ Correct Option: 1

We know the log property: log⁡b(xy)=log⁡bx+log⁡by\log_b(xy) = \log_b x + \log_b y

So we can expand the first term:

2log⁡3(3n)=2[log⁡33+log⁡3n]=2[1+log⁡3n]=2+2log⁡3n2\log_3(3n) = 2[\log_3 3 + \log_3 n] = 2[1 + \log_3 n] = 2 + 2\log_3 n

The full expression becomes:

2+2log⁡3n−log⁡3(n2+1)2 + 2\log_3 n - \log_3(n^2 + 1)


Now, when nn is sufficiently large, n2+1≈n2n^2 + 1 \approx n^2 because the +1+1 becomes negligible compared to n2n^2.

So we replace:

log⁡3(n2+1)≈log⁡3(n2)\log_3(n^2 + 1) \approx \log_3(n^2)

Using the power rule log⁡b(xk)=k⋅log⁡bx\log_b(x^k) = k \cdot \log_b x:

log⁡3(n2)=2log⁡3n\log_3(n^2) = 2\log_3 n


Substituting back into the expression:

2+2log⁡3n−2log⁡3n2 + 2\log_3 n - 2\log_3 n

=2= 2


The approximate value is 2\boxed{2}

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