How do you write Log(base 2)36 in the form of A and B if A=log (base 2) 2 and B=log(base 2) 3 ??

Question

I really don't get the question.. Can anyone help me please!?

Answer 1

log(base2)36
= log(base2)[2.2.3.3]
= log(base2) [2.2] + log(base2) [3.3]
= 2log(base2)2 + 2log(base2)3
= 2A + 2B
= 2(A+B)

Answer 2

Well, lets keep it simple. 36 is 2*2*3*3, right? So Log 36 = Log (2*2*3*3)......and this can be seperated to Log (2*2) + Log (3*3), which is = 2Log 2 + 2Log 3= 2A + 2B. Got the hang of it??

Answer 3

Here you are!

For simplicity log will denote log(base2) ok?
A = log 2
B = log 3

log 36 = log 6^2 = 2 log 6 = 2 log 2*3 = 2 (log 2 + log 3) = 2A + 2B

or

log 36 = 2 * (1 + log 3)

Answer 4

You need to break down 36 into primes, and use some log rules.

36=2*2*3*3. So we have log(base2) (2*2*3*3). Using log rules, we can make this log(2) (2*2)+log(2) (3*3), and then 2log(2) (2)+2log(2) (3). This is 2A+2B.

Answer 5

Log (base 2) 36 = log (base 2) 6^2
= 2log(base 2) 6 = 2log (base 2) 2*3
=2[log(base 2) 2 +log(base 2) 3]
= 2(A + B)

This solution uses two important laws of logarithms:

log (base y) z ^n = nlog (base y) z, and
log (base y) a*b = log(base y) a +log(base y) b

Answer 6

let me simplify the word log(base 2) be logx.
so, log(base 2)36 = logx 36 = logx (2 *18) = logx 2 + logx 18 = logx 2 + logx 2 + logx 9 = 2 * (logx 2) + logx 9 = 2 * (logx 2) + 2 * (logx 3)