Clear explanations would be appreciated as I just started out on this type of Mathematics and I'm having a bad time with it.
2007-02-20 19:31:51 · 5 answers · asked by tempo_tranquillo 2 in Science & Mathematics ➔ Mathematics
lg = log base 10
2007-02-20 19:57:53 · update #1
Hmmm....
lg50 = lg(2*5*5) = lg2 + 2*lg5
so lg2lg50 = (lg2)^2 + 2lg2lg5
so (lg5)^2 + lg2lg50 =
(lg5)^2 + (lg2)^2 + 2lg2lg5 =
(lg5 + lg2)^2 =
(lg10)^2
Well if lg is the log base 10, the the answer is just 1^2 = 1, otherwise, I am not sure how to evaluate this....
2007-02-20 19:43:09 · answer #1 · answered by Phineas Bogg 6 · 1⤊ 0⤋
I assume all the logs are base 10.
(log 5)² + (log 2)(log 50)
= (log 10/2)² + (log 2)(log 100/2)
= [(log 10) - (log 2)]² + (log 2)[(log 100) - (log 2)]
= [1 - (log 2)]² + (log 2)[2 - (log 2)]
= 1 - 2(log 2) + (log 2)² + 2(log 2) - (log 2)²
= 1
2007-02-20 20:01:42 · answer #2 · answered by Northstar 7 · 1⤊ 0⤋
is this log 10 or log 2?
2007-02-20 19:54:54 · answer #3 · answered by Experimental876 4 · 0⤊ 0⤋
LHS=(lg5)^2+lg2*[lg(5*5*2)]
=(lg5)^2+lg2*(lg5+lg5+lg2)
=(lg5)^2+lg2*(2lg5+lg2)
=(lg5)*2+2*lg2*lg5+(lg2)^2
Leave the lg5 as (a) and lg2 as (b)
then
LHS=a^2+2*a*b+b^2=(a+b)^2
=(lg5+lg2)^2=(lg10)^2=1^2=1
answer = 1
2007-02-20 19:45:54 · answer #4 · answered by QuizBox 2 · 0⤊ 0⤋
nerd
2007-02-20 19:38:50 · answer #5 · answered by latino till the d 1 · 0⤊ 2⤋