Question

Log_2⁡56 +2log_2⁡12-log_2⁡63

Answer 1

$log_{2}56+2log _{2}12-log _{2}63=log _{2}56+log _{2}144-log _{2}63=log _{2} \frac{56*144}{63} =log_{2}128=log _{2} 2^{7}=7log _{2}2=7$

Answer 2

To solve the expression, we'll use the properties of logarithms. The three main properties we'll apply are:

1. log_b(x) + log_b(y) = log_b(xy)
2. log_b(x) - log_b(y) = log_b(x/y)
3. log_b(x^n) = n * log_b(x)

Now, let's simplify the expression step by step:

1. log_2(56) + 2 * log_2(12) - log_2(63)

First, we can simplify each term:

• log_2(56) can be expressed as log_2(8 * 7) = log_2(2^3 * 7) = log_2(2^3) + log_2(7) = 3 + log_2(7).
• log_2(12) can be expressed as log_2(4 * 3) = log_2(2^2 * 3) = log_2(2^2) + log_2(3) = 2 + log_2(3).
• log_2(63) can be expressed as log_2(9 * 7) = log_2(3^2 * 7) = log_2(3^2) + log_2(7) = 2 + log_2(7).

Now, our expression becomes:

= (3 + log_2(7)) + 2 * (2 + log_2(3)) - (2 + log_2(7))

Next, we can combine like terms:

= 3 + log_2(7) + 4 + 2 * log_2(3) - 2 - log_2(7)

Now, the log_2(7) terms cancel each other out:

= 3 + 4 + 2 * log_2(3) - 2

Now, we have:

= 7 + 2 * log_2(3)

So, the simplified expression is 7 + 2 * log_2(3).