### Exponential and logarithmic functions - Examples

• Without using a calculator, find the values of the following expressions.
(a) $$\frac{{\log 32 + \log 0.25}}{{\log 5 - 5\log 2 - 1}}$$

(b) $$\frac{{2 + \log 16 - 2\log 0.04}}{{4\log \sqrt 5 - \log 0.25}}$$
• $$\begin{array}{1}(a)\frac{{\log 32 + \log 0.25}}{{\log 5 - 5\log 2 - 1}} = \frac{{\log 32 + \log 0.25}}{{\log 5 - \log {2^5} - \log 10}}\\ = {\frac{{\log 32 + \log 0.25}}{{\log 5 - \log 32 - \log 10}}^{}}\\ = {\frac{{\log (32 \times 0.25)}}{{\log {\textstyle{5 \over {32{\rm{ }} \times {\rm{ }}10}}}}}^{}}\\ = {\frac{{\log 8}}{{\log {\textstyle{1 \over {64}}}}}^{}}\\ = {\frac{{\log 8}}{{\log {8^{ - 2}}}}^{}}\\ = {\frac{{\log 8}}{{ - {\rm{ }}2\log 8}}^{}}\\ = {\underline{\underline { - {\rm{ }}\frac{1}{2}}} ^{}}\end{array}$$

$$\begin{array}{1}(b)\frac{{2 + \log 16 - 2\log 0.04}}{{4\log \sqrt 5 - \log 0.25}} = \frac{{\log 100 + \log 16 - \log {{({\textstyle{4 \over {100}}})}^2}}}{{\log {5^2} - \log {\textstyle{{25} \over {100}}}}}\\ = {\frac{{\log {\textstyle{{100{\rm{ }} \times {\rm{ }}16} \over {{\textstyle{{16} \over {10{\rm{ }}000}}}}}}}}{{\log {\textstyle{{25} \over {{\textstyle{{25} \over {100}}}}}}}}^{}}\\ = {\frac{{\log 1{\rm{ }}000{\rm{ }}000}}{{\log 100}}^{^{}}}\\ = \frac{6}{2}\\ = {\underline{\underline 3} ^{}}\end{array}$$
• If α and β are the two roots of the equation $$2{x^2} - 50x + 20 = 0$$, find the value of $$\log \alpha + \log \beta$$.
• $$\begin{array}{1}2{x^2} - 50x + 20 = 0\\\alpha \beta = Product Of Roots\\ = \frac{{20}}{2}\\ = 10\end{array}$$
∴ $$\begin{array}{1}\log \alpha + \log \beta = \log \alpha \beta \\ = \log 10\\ = \underline{\underline {{\rm{ }}1{\rm{ }}}} \end{array}$$
• Solve the equation $$\log (3{x^2} - 16) = \log (2x - 11)$$.
• $$\log (3{x^2} - 16) = \log (2x - 11)$$
Assume that $$3{x^2} - 16 > 0$$ and $$2x - 11 > 0$$.
∴ $$\begin{array}{c}3{x^2} - 16 = 2x - 11\\3{x^2} - 2x - 5 = 0\\(x + 1)(3x - 5) = {0^{}}\end{array}$$
$$x = - 1$$ or $$x = \frac{5}{3}$$
When $$x = - 1$$,
$$\begin{array}{c}3{x^2} - 16 = 3{( - 1)^2} - 16\\ = 3 - {16^{}}\\ = - {13^{}}\\ < {0^{}}\end{array}$$ and $$\begin{array}{c}2x - 11 = 2( - 1) - {11^{}}\\ = - 2 - {11^{}}\\ = - {13^{}}\\ < {0^{}}\end{array}$$
When $$x = \frac{5}{3}$$,
$$\begin{array}{c}3{x^2} - 16 = 3{(\frac{5}{3})^2} - 16\\ = \frac{{25}}{3} - {16^{^{^{}}}}\\ = - \frac{{23}}{3}\\ < 0\end{array}$$ and $$\begin{array}{c}2x - 11 = 2(\frac{5}{3}) - 11\\ = \frac{{10}}{3} - {11^{^{^{}}}}\\ = - \frac{{23}}{3}\\ < 0\end{array}$$
∵ When $$x = - 1$$ and $$x = \frac{5}{3}$$, both $$\log (3{x^2} - 16)$$ and $$\log (2x - 11)$$ are undefined.