Writing expressions as a Single logarithm
Write each of the following as a single logarithm.
(a) \log_{a}7\,+\,4\log_{a}3\, (b) {\frac{2}{3}}\ln8-\ln\left({{5}}^{2}-1\right)
(c) \log_{a}x\,+\,\log_{a}9\,+\,\log_{a}(x^{2}\,+\,1\,){-}\,\log_{a}5
(a) \log_{a}7\,+\,4\log_{a}3\,=\,\log_{a}7\,+\,\log_{a}3^{4} r\log_{a}M=\log_{a}M^{r}
=\log_{a}7\ +\log_{a}81=\log_{a}(7\cdot81) \log_{a}M+\log_{a}N=\textstyle\log_{a}(M\cdot N)
=\log_{a}567(b) {\frac{2}{3}}\mathrm{ln}\,8-\,\mathrm{ln}\left(5^{2}-1\right)\,=\,\mathrm{ln}\,8^{2/3}\,-\,\mathrm{ln}\left(25\,-\,1\right) r\log_{a}M=\log_{a}M^{r}
=\,\ln 4 – \ln 24 8^{2/3}=(~{\sqrt[3]{8}})^{2}=2^{2}=4
=\ln\!\left({\frac{4}{24}}\right) \log_{a}M-\log_{a}N=\log_{a}\!\left(\frac{M}{N}\right)
{{=\ln\!\left({\frac{1}{6}}\right)}}{{=\,\ln1\,-\,\ln6}} \\ =-\ln6 \ln 1=0
(c) \log_{a}x\,+\,\log_{a}9\,+\,\log_{a}(x^{2}\,+\,1\,)\,-\,\log_{a}5 = \log_{a}(9x)+\log_{a}(x^{2}+1)-\log_{a}5
=\log_{a}[9x(x^{2}+1)\;]\ -\log_{a}5 \\=\log_{a}\!\left[{\frac{9x(x^{2}+1)}{5}}\right]