Question

Value of: $2log_62 + 3log_63 + log_612$ is

• A. 2
• B. 3
• C. 11
• D. 10

Answer 1

4

Answer 2

Let the given expression be $E$.

$E = 2log_62 + 3log_63 + log_612$

We use the properties of logarithms:

1. $a \log_b x = \log_b x^a$
2. $\log_b x + \log_b y = \log_b (xy)$
3. $\log_b b = 1$

Apply property 1 to the first two terms:

$2log_62 = log_6 2^2 = log_6 4$

$3log_63 = log_6 3^3 = log_6 27$

Substitute these back into the expression:

$E = log_6 4 + log_6 27 + log_6 12$

Apply property 2 to combine the terms:

$E = log_6 (4 \times 27 \times 12)$

Calculate the product inside the logarithm:

$4 \times 27 \times 12 = 108 \times 12 = 1296$

So the expression becomes:

$E = log_6 1296$

Now we need to evaluate $log_6 1296$. This means finding the power to which 6 must be raised to get 1296.

Let $6^x = 1296$.

We can calculate the powers of 6:

$6^1 = 6$

$6^2 = 36$

$6^3 = 216$

$6^4 = 1296$

So, $6^4 = 1296$.

Therefore, $log_6 1296 = 4$.