SOLUTION: Hi, I have been going over this more time then you know. I need to use the laws of logarithms to express logarithm in base 2 of 6 minus logarithm in base 2 of 3 plus 2logarithm in

Question 28781: Hi, I have been going over this more time then you know. I need to use the laws of logarithms to express logarithm in base 2 of 6 minus logarithm in base 2 of 3 plus 2logarithm in base of 2 sqrt 8.
This is what I have,
Original equation
= logarithm in base 2 of 6 - log in base of 2 of 3 + log in base of 2 of 8^1/2
= SAME + log in base of 2 of 2^3 (base of two and 2 to the power of cancel out)
=Same as top + log3
=log in base of 2(6)(3) - log3
=log in base of 2 = (18/3)
log2 = 6
I really hope this makes sense, because I'd really appreciate help. I'm not sure if I did this right, it kind of seems like I haven't so thats why I'm here double checking. Thank you for your time.
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
all my log terms here are to base 2, i just omit that in the writing, to make it clearer...

log6 - log3 + log8^(1/2)
log(2*3) - log3 + (1/2)log8
log(2*3) - log3 + (1/2)log2^3
log(2*3) - log3 + (3/2)log2
log(2) + log(3) - log3 + (3/2)log2

now log[base2](2) = 1

so, we get
1 + log(3) - log3 + (3/2)(1)
1 + (3/2)
5/2

jon.

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