|dw:1333255112352:dw| can you take it from here?
hint: log2(1) = 0 log2(2) = 1 log2(3) = 1.584962500721156 log2(4) = 2 log2(5) = 2.321928094887362 log2(6) = 2.584962500721156 log2(7) = 2.807354922057604 log2(8) = 3 log2(9) = 3.169925001442312 log2(10) = 3.321928094887362 Note that log2(x) is defined for any x greater than zero. If you have a calculator than computes the natural logarithm (often denoted ln), then you can calculate log2(x) = ln(x)/ln(2). The same thing works with log base 10, i.e. log2(x) = log10(x)/log10(2).
dpalnc_ I thought that the log base 2 cancels out and then the equation is: a^2-6a= 10+3a? Then you take it from there.
silly me... yeah, i should've done that.... but the equation will still be the same... I like to do it the long way... NOT! :)
can you take it from there then?
the answer is -1, 10...but I can't figure out how to get there
yes, a = 10 , a = -1
it's probably the quadratic equation then...
Never mind!!! I figured it out...thanks!!!
no, it's factoring
|dw:1333255705473:dw| factor that left side...
didn't read the above post that you figured it out already...
your right! I was wrong with the +10...it should have been -10