23) log(little3)(2x)^2 exand this
30) log(little2)9 use the change of base formula to solve
44) 1/2log(little2)15-log(little5)sqrt75 use the properties of logarithims to evaluate
23)
log3(2x)2=2log3(2x)=2(log3(2)+log3(x))=2log3(2)+2log3(x)
30)
log29=log9log2 And you can finish it with a calculator.
44)
12log215−log5√75=12log2(5∗3)−log5(25∗3)1/2=log2(5∗3)1/2−log5(25∗3)1/2=log2(51/2∗31/2)−log5(251/2∗31/2)=log2(51/2∗31/2)−[log5(5∗31/2)]=log251/2+log231/2−[log55+log531/2]=log251/2+log231/2−[1+log531/2]=log251/2+log231/2−22−12log53=12[log25+log23−2−log53]
I was working this out as I was writing the code so it is probably not the best way.
Plus i am not very happy with the answer, but this is as far as I get.
Maybe someone esle will do something more/different. (Are you sure the base (little) numbers were correct?)
23)
log3(2x)2=2log3(2x)=2(log3(2)+log3(x))=2log3(2)+2log3(x)
30)
log29=log9log2 And you can finish it with a calculator.
44)
12log215−log5√75=12log2(5∗3)−log5(25∗3)1/2=log2(5∗3)1/2−log5(25∗3)1/2=log2(51/2∗31/2)−log5(251/2∗31/2)=log2(51/2∗31/2)−[log5(5∗31/2)]=log251/2+log231/2−[log55+log531/2]=log251/2+log231/2−[1+log531/2]=log251/2+log231/2−22−12log53=12[log25+log23−2−log53]
I was working this out as I was writing the code so it is probably not the best way.
Plus i am not very happy with the answer, but this is as far as I get.
Maybe someone esle will do something more/different. (Are you sure the base (little) numbers were correct?)