SOLUTION: This one looked easy but it's got me confused please help and thank you. Use the laws of logarithms to express log base2 6 - log base 2 3 + 2 log base 2 sqrt8 as a single logarith

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: This one looked easy but it's got me confused please help and thank you. Use the laws of logarithms to express log base2 6 - log base 2 3 + 2 log base 2 sqrt8 as a single logarith      Log On


   

Question 280462: This one looked easy but it's got me confused please help and thank you.
Use the laws of logarithms to express log base2 6 - log base 2 3 + 2 log base 2 sqrt8 as a single logarithm then evaluate. I got this far log base 2^6 - log base2^3 +2 log base 2 sqrt 8 =log base 2 6/3 sqrt8^2??? help
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C6%29+-+log%282%2C3%29+%2B+2%2Alog%282%2Csqrt%288%29%29
The 1st 2 terms are:
log%282%2C6%29+-+log%282%2C3%29+=+log%282%2C6%2F3%29
log%282%2C6%2F3%29+=+log%282%2C2%29
log%282%2C2%29+=+1
So far I have
1+%2B+2%2Alog%282%2Csqrt%288%29%29+=+1+%2B+log%282%2C%28sqrt%288%29%29%5E2%29
1+%2B+log%282%2C%28sqrt%288%29%29%5E2%29+=+1+%2B+log%282%2C8%29
1+%2B+log%282%2C8%29+=+1+%2B+3
1+%2B+3+=+4