Bab 5 Indeks dan Logaritma

$\begin{array}{l}lo{g}_{16}\left[lo{g}_2\left(5x–4\right)\right]=lo{g}_9\sqrt{3}\\ lo{g}_{16}\left[lo{g}_2\left(5x–4\right)\right]=\frac{1}{4}\leftarrow \boxed{\begin{array}{l}{\log }_9\sqrt{3}={\log }_9{3}^{\frac{1}{2}}=\frac{1}{2}{\log }_{9}3\\ =\frac{1}{2}\left(\frac{1}{{\log }_{3}9}\right)=\frac{1}{2}\left(\frac{1}{2}\right)=\frac{1}{4}\end{array}}\\ lo{g}_2\left(5x–4\right)={16}^{\frac{1}{4}}\\ lo{g}_2\left(5x–4\right)=2\\ 5x–4=2^2\\ 5x=8\\ x=\frac{8}{5}\end{array}$

$\begin{array}{l}{5}^{{\log }_{4}x}=125\\ {\log }_5{5}^{{\log }_{4}x}={\log }_{5}125\leftarrow \boxed{\begin{array}{l}\text{ambil log asas 5}\\ \text{di kedua-dua belah}\end{array}}\\ \left({\log }_{4}x\right)\left({\log }_{5}5\right)=3\\ \left({\log }_{4}x\right)\left(1\right)=3\\ x=4^3=64\end{array}$

$\begin{array}{l}{5}^{{\log }_5\left(x+1\right)}=9\\ {\log }_5{5}^{{}^{{\log }_5\left(x+1\right)}}={\log }_{5}9\\ {\log }_5\left(x+1\right).{\log }_{5}5={\log }_{5}9\\ {\log }_5\left(x+1\right)={\log }_{5}9\\ x+1=9\\ x=8\end{array}$