Question

could someone please help me with this question?
i got 0.5.. but the back of my book says -0.5
im not sure if it's a typo, or if i did it wrong?

\[\log_{2}\sqrt{36} - \log_{2} \sqrt{72}\]

Answer 1

\[\log_{2}\sqrt{36} - \log_{2} \sqrt{72}\]


\[\log_{2}\left(\frac{\sqrt{36}}{\sqrt{72}}\right)\]


\[\log_{2}\left(\sqrt{\frac{36}{72}}\right)\]


See where to go from here?

Answer 2

equate √(36/72) ?

Answer 3

\[\log_{2}\sqrt{36} - \log_{2} \sqrt{72}\]


\[\log_{2}\left(\frac{\sqrt{36}}{\sqrt{72}}\right)\]


\[\log_{2}\left(\sqrt{\frac{36}{72}}\right)\]


\[\log_{2}\left(\sqrt{\frac{1}{2}}\right)\]


\[\log_{2}\left(\frac{\sqrt{1}}{\sqrt{2}}\right)\]


\[\log_{2}\left(\frac{1}{\sqrt{2}}\right)\]


\[\log_{2}\left(\frac{1}{2^{\frac{1}{2}}}\right)\]


\[\log_{2}\left(2^{-\frac{1}{2}}\right)\]


How about now?

Answer 4

u could convert them to base 10 and then calculate....

Answer 5

uh... how do you do that? :S

Answer 6

\[\log_{2}\sqrt{36} - \log_{2} \sqrt{72}\]


\[\log_{2}\left(\frac{\sqrt{36}}{\sqrt{72}}\right)\]


\[\log_{2}\left(\sqrt{\frac{36}{72}}\right)\]


\[\log_{2}\left(\sqrt{\frac{1}{2}}\right)\]


\[\log_{2}\left(\frac{\sqrt{1}}{\sqrt{2}}\right)\]


\[\log_{2}\left(\frac{1}{\sqrt{2}}\right)\]


\[\log_{2}\left(\frac{1}{2^{\frac{1}{2}}}\right)\]


\[\log_{2}\left(2^{-\frac{1}{2}}\right)\]


\[-\frac{1}{2}\log_{2}\left(2\right)\]


\[-\frac{1}{2}(1)\]


\[-\frac{1}{2}\]


So

\[\log_{2}\sqrt{36} - \log_{2} \sqrt{72} = -\frac{1}{2}\]

Answer 7

I'm using the idea that

\[\Large \log_{b}(X)-\log_{b}(Y)=\log_{b}\left(\frac{X}{Y}\right)\]

Answer 8

Also,

\[\Large \log_{b}(x^{y})= y\log_{b}(x)\]

Answer 9

convert to base 10 as follows :