+118724 log(sqrt(27))/log(1*2)+log(8)/log(1*2)-log(sqrt(1000))/log(1*2)
\(\frac{log(\sqrt{27})}{log(1*2)}+\frac{log(8)}{log(1*2)}-\frac{log(\sqrt{1000})}{log(1*2)}\\ =\frac{log(27)^{0.5}}{log(2)}+\frac{log(2^3)}{log(2)}-\frac{log(1000)^{0.5}}{log(2)}\\ =\frac{log(3^3)^{0.5}}{log(2)}+\frac{log(2^3)}{log(2)}-\frac{log(10^3)^{0.5}}{log(2)}\\ =\frac{log(3)^{1.5}}{log(2)}+\frac{log(2^3)}{log(2)}-\frac{log(10)^{1.5}}{log(2)}\\ =\frac{log(3)^{1.5}+log(2^3)-log(10)^{1.5}}{log(2)}\\ =\frac{log(3)^{1.5}+log(2^3)-1.5log(10)}{log(2)}\\ =\frac{1.5log(3)+3log(2)-1.5}{log(2)}\\ \)
I am really not sure how it would be best to present this. :/
Plus I have assumed that this is base 10.
What about if it is not base 10.....
then.
\(=\frac{log(3)^{1.5}+log(2^3)-log(10)^{1.5}}{log(2)}\\ =\frac{log(3^{1.5}\div 10^{1.5})+log(2^3)}{log(2)}\\ =\frac {log(0.3^{1.5})+log(4^{0.5})^3} {log(2)} \\ =\frac {log(0.3^{1.5})+log(4^{1.5})} {log(2)} \\ =\frac {log(0.3^{1.5}*4^{1.5})} {log(2)} \\ =\frac {log(1.2^{1.5})} {log(2)} \\ =\frac {1.5log(1.2)} {log(2)} \\ \)
