**5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)**

**Soalan 1:**

Selesaikan persamaan, log_{3} [log_{2}(2*x* – 1)] = 2

*Penyelesaian:*

log_{3} [log_{2} (2*x* – 1)] = 2 ← (jika log _{a}_{ }*N* = *x*, *N* = *a*^{x})

log_{2} (2*x* – 1) = 3^{2}

log_{2} (2*x* – 1) = 9

2*x* – 1 = 2^{9}

*x ***= 256.5**

**Soalan 2**

Selesaikan persamaan, $lo{g}_{16}\left[lo{g}_{2}\left(5x\u2013\text{}4\right)\right]=lo{g}_{9}\sqrt{3}$

*Penyelesaian:*

$\begin{array}{l}lo{g}_{16}\left[lo{g}_{2}\left(5x\u2013\text{}4\right)\right]=lo{g}_{9}\sqrt{3}\\ lo{g}_{16}\left[lo{g}_{2}\left(5x\u2013\text{}4\right)\right]=\frac{1}{4}\leftarrow \overline{)\begin{array}{l}{\mathrm{log}}_{9}\sqrt{3}={\mathrm{log}}_{9}{3}^{\frac{1}{2}}=\frac{1}{2}{\mathrm{log}}_{9}3\\ =\frac{1}{2}\left(\frac{1}{{\mathrm{log}}_{3}9}\right)=\frac{1}{2}\left(\frac{1}{2}\right)=\frac{1}{4}\end{array}}\\ lo{g}_{2}\left(5x\u2013\text{}4\right)={16}^{\frac{1}{4}}\\ lo{g}_{2}\left(5x\u2013\text{}4\right)=2\\ 5x\u2013\text{}4={2}^{2}\\ 5x=8\\ x=\frac{8}{5}\end{array}$
**Soalan 3**

Selesaikan persamaan, ${5}^{{\mathrm{log}}_{4}x}=125$

*Penyelesaian:*

$\begin{array}{l}{5}^{{\mathrm{log}}_{4}x}=125\\ {\mathrm{log}}_{5}{5}^{{\mathrm{log}}_{4}x}={\mathrm{log}}_{5}125\leftarrow \overline{)\begin{array}{l}\text{ambillogasas5}\\ \text{dikedua-duabelah}\end{array}}\\ \left({\mathrm{log}}_{4}x\right)\left({\mathrm{log}}_{5}5\right)=3\\ \left({\mathrm{log}}_{4}x\right)\left(1\right)=3\\ x={4}^{3}=64\end{array}$
**Soalan 4**

Selesaikan persamaan, ${5}^{{\mathrm{log}}_{5}\left(x+1\right)}=9$

*Penyelesaian:*

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