Exercise
\frac{1}{2}+\log_9 {\sqrt{18}}-\log_9 {\frac{1}{\sqrt{2}}}+\log_9 4.5=
Final Answer
Solution
Using Logarithm Rules we get:
\frac{1}{2}+\log_9 {\sqrt{18}}-\log_9 {\frac{1}{\sqrt{2}}}+\log_9 4.5=
=\frac{1}{2}+\log_9 {\sqrt{9}\sqrt{2}}-\log_9 {\sqrt{2}^{-1}}+\log_9 4.5=
=\frac{1}{2}+\log_9 {3\sqrt{2}}-\log_9 {\sqrt{2}^{-1}}+\log_9 4.5=
=\frac{1}{2}+\log_9 3 +\log_9 \sqrt{2}-\log_9 {\sqrt{2}^{-1}}+\log_9 4.5=
=\frac{1}{2}+\log_9 3 +\log_9 \sqrt{2}-(-1)\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\log_9 3 +\log_9 \sqrt{2}+\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\frac{\log_3 3}{\log_3 9} +\log_9 \sqrt{2}+\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\frac{\log_3 3}{\log_3 3^2} +\log_9 \sqrt{2}+\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\frac{\log_3 3}{2\log_3 3} +\log_9 \sqrt{2}+\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\frac{1}{2} +\log_9 \sqrt{2}+\log_9 {\sqrt{2}}+\log_9 4.5=
=\frac{1}{2}+\frac{1}{2} +\log_9 {(\sqrt{2}\sqrt{2})}+\log_9 4.5=
=\frac{1}{2}+\frac{1}{2} +\log_9 {2}+\log_9 4.5=
=1 +\log_9 {2}+\log_9 4.5=
=1 +\log_9 {2}+\log_9 \frac{9}{2}=
=1 +\log_9 {2}+\log_9 9 - \log_9 2=
=1 +\log_9 {2}+1 - \log_9 2=
=1 +1 =2