The approximate value of the expression 2_3 3n - _3(n^2 + 1) for a sufficiently large n is Option 1: 2 Option 2: 3 Option 3: 2 - _3 2 Option 4: 1

Answer: 2. Solution: We know the log property: _b(xy) = _b x + _b y So we can expand the first term: 2_3(3n) = 2[_3 3 + _3 n] = 2[1 + _3 n] = 2 + 2_3 n The full expression becomes: 2 + 2_3 n - _3(n^2 + 1) Now, when n is sufficiently large, n^2 + 1 n^2 because the +1 becomes negligible compared to n^2. So we replace: _3(n^2 + 1) _3(n^2) Using the power rule _b(x^k) = k * _b x: _3(n^2) = 2_3 n Substituting back into the expression: 2 + 2_3 n - 2_3 n = 2 The approximate value is 2

Modern Math

Logarithms

Easy

The approximate value of the expression $2\log_3 3n - \log_3(n^2 + 1)$ for a sufficiently large $n$ is

$2 - \log_3 2$

✅ Correct Option: 1

We know the log property: $\log_b(xy) = \log_b x + \log_b y$

So we can expand the first term:

$2\log_3(3n) = 2[\log_3 3 + \log_3 n] = 2[1 + \log_3 n] = 2 + 2\log_3 n$

The full expression becomes:

$2 + 2\log_3 n - \log_3(n^2 + 1)$

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

So we replace:

$\log_3(n^2 + 1) \approx \log_3(n^2)$

Using the power rule $\log_b(x^k) = k \cdot \log_b x$ :

$\log_3(n^2) = 2\log_3 n$

Substituting back into the expression:

$2 + 2\log_3 n - 2\log_3 n$

$= 2$

The approximate value is $\boxed{2}$

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