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)
$$log_3(2x)^2\\
=2log_3(2x)\\
=2(log_3(2)+log_3(x)) \\
=2log_3(2)+2log_3(x) \\$$
30)
$$log_29=\frac{log9}{log2}$$ And you can finish it with a calculator.
44)
$$\frac{1}{2}log_215-log_5\sqrt{75}\\\\
=\frac{1}{2}log_2(5*3)-log_5(25*3)^{1/2}\\\\
=log_2(5*3)^{1/2}-log_5(25*3)^{1/2}\\\\
=log_2(5^{1/2}*3^{1/2})-log_5(25^{1/2}*3^{1/2})\\\\
=log_2(5^{1/2}*3^{1/2})-[log_5(5*3^{1/2})]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-[log_5\;5+log_53^{1/2}]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-[1+log_53^{1/2}]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-\frac{2}{2}-\frac{1}{2}log_53\\\\
=\frac{1}{2} [log_2\;5+log_2\;3-2-log_53]\\\\$$
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)
$$log_3(2x)^2\\
=2log_3(2x)\\
=2(log_3(2)+log_3(x)) \\
=2log_3(2)+2log_3(x) \\$$
30)
$$log_29=\frac{log9}{log2}$$ And you can finish it with a calculator.
44)
$$\frac{1}{2}log_215-log_5\sqrt{75}\\\\
=\frac{1}{2}log_2(5*3)-log_5(25*3)^{1/2}\\\\
=log_2(5*3)^{1/2}-log_5(25*3)^{1/2}\\\\
=log_2(5^{1/2}*3^{1/2})-log_5(25^{1/2}*3^{1/2})\\\\
=log_2(5^{1/2}*3^{1/2})-[log_5(5*3^{1/2})]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-[log_5\;5+log_53^{1/2}]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-[1+log_53^{1/2}]\\\\
=log_2\;5^{1/2}+log_2\;3^{1/2}-\frac{2}{2}-\frac{1}{2}log_53\\\\
=\frac{1}{2} [log_2\;5+log_2\;3-2-log_53]\\\\$$
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?)