Mathematics - Basic maths Question with Solution | TestHub

$$\text{ A) }{\log }_{3}5=x\Rightarrow 3^x=5\Rightarrow 5^{1/x}=3.\dots \dots \dots \left(1\right)$$

and ${\log }_{25}11=\frac{{\log }_{5}11}{2}=\mathrm{y}$

$\therefore {\log }_{5}11=2\mathrm{y}\Rightarrow 5^{2\mathrm{y}}=11$..............(2)

$(2÷1)\Rightarrow \frac{11}{5}=5^{2y-\frac{1}{x}}=5\frac{2xy-1}{x}$ ${\log }_3\left(\frac{11}{3}\right)=\frac{2\mathrm{xy}-1}{\mathrm{x}}{\log }_{3}5=\frac{2\mathrm{xy}-1}{\mathrm{x}}\cdot \mathrm{x}=2\mathrm{xy}-1$.

B) We have $(2x{)}^{{\log }_{5}2}-(3x{)}^{{\log }_{5}3}$ ∴ Taking logarithm to the base 5 on both sides, we get $\left({\log }_{5}2\right)\cdot \left({\log }_{5}2+{\log }_5\mathrm{x}\right)=\left({\log }_{5}3\right)\cdot \left({\log }_{5}3+{\log }_5\mathrm{x}\right)$ $\Rightarrow -\left({\log }_{5}3-{\log }_{5}2\right)\cdot {\log }_5\mathrm{x}=\left({\log }_{5}3-{\log }_{5}2\right)\cdot \left({\log }_{5}3+{\log }_{5}2\right)$ $\Rightarrow {\log }_5\left(\frac{1}{\mathrm{x}}\right)={\log }_{5}6\Rightarrow \mathrm{x}=\frac{1}{6}\equiv {\mathrm{x}}_0$ (Given) Hence, $\left({\mathrm{x}}_0+\frac{1}{{\mathrm{x}}_0}\right)=\frac{37}{6}$

C) ${\log }_{\frac{1}{\mathrm{c}}}\sqrt{2.4}=\frac{-1}{2}{\log }_{\mathrm{c}}\frac{12}{5}=\frac{1}{2}\left[{\log }_{\mathrm{c}}5-{\log }_{\mathrm{c}}12\right]=\frac{1}{2}\left[{\log }_{\mathrm{c}}5-\left(2{\log }_{\mathrm{c}}2+{\log }_{\mathrm{c}}3\right)\right]$ $=\frac{1}{2}[\mathrm{z}-2\mathrm{w}-\mathrm{y})=\frac{1}{2}(\mathrm{z}-\mathrm{y}-2\mathrm{w})]$

D) $9^x=15\Rightarrow 3^{2x}=15\Rightarrow 2x{\log }_{10}3={\log }_{10}(3\times 5)$ $2\mathrm{x}.\mathrm{b}={\log }_{10}3+{\log }_{10}\left(\frac{10}{2}\right)=\mathrm{b}+1-\mathrm{a}\therefore \mathrm{x}=\frac{1+\mathrm{b}-\mathrm{a}}{2\mathrm{b}}$.