Surds And Indices

$$\text{1}{\text{0}}^{-8x}$$
$$={\left[{10}^x\right]}^{-8}$$
$$={\left[\frac{1}{2}\right]}^{-8}$$
$$=2^8$$
$$=256$$

$$\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\right)$$
Formula :
$$\frac{1}{\text{Difference denominator value}}\times$$     $$\left[\frac{1}{\text{First value}}-\frac{1}{\text{Last value}}\right]$$
$$=\frac{1}{3}\times$$ $$\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right]$$
$$=\frac{1}{3}\times \left[1-\frac{1}{16}\right]$$
$$=\frac{1}{3}\times \frac{15}{16}$$
$$=\frac{5}{16}$$

$$\frac{1}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$
$$=\frac{\sqrt{5}-\sqrt{3}}{5-3}$$
$$=\frac{2.236-1.732}{2}$$
$$=\frac{0.504}{2}$$
$$=0.252$$

$${\left(\frac{3}{5}\right)}^3{\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$$
$$\Rightarrow {\left(\frac{3}{5}\right)}^{\left(3-6\right)}={\left(\frac{3}{5}\right)}^{2x-1}$$
$$\Rightarrow {\left(\frac{3}{5}\right)}^{-3}={\left(\frac{3}{5}\right)}^{2x-1}$$
$$\Rightarrow 2x-1=-3$$
$$\Rightarrow 2x=-2$$
$$\Rightarrow x=-1$$

$$\text{5}\sqrt{5}\times 5^3÷5^{-\frac{3}{2}}\text{ = }{\text{5}}^{a+2}$$
$$\Rightarrow \frac{5\times 5^{\frac{1}{2}}\times 5^3}{5^{-\frac{3}{2}}}=5^{a+2}$$
$$\Rightarrow 5^{\left(1+\frac{1}{2}+3+\frac{3}{2}\right)}=5^{a+2}$$
$$\Rightarrow 5^6=5^{a+2}$$
$$\Rightarrow a+2=6$$
$$\Rightarrow a=4$$

$${\text{2}}^x=\sqrt[3]{32}$$
$$\Rightarrow {\text{2}}^x={\left(32\right)}^{\frac{1}{3}}={\left(2^5\right)}^{\frac{1}{3}}$$
$$\Rightarrow x=\frac{5}{3}$$

$$\sqrt{12+\sqrt{12+\sqrt{12+.....}}}$$
(3, 4) are the factor of 4
If there is '+' in $$\sqrt{}$$  Answer is Highest value
If there is '-' in $$\sqrt{}$$  Answer is lowest value.
So the answer is 4

$${\text{5}}^{\left(x+3\right)}\text{ = 2}{\text{5}}^{(3x-4)}$$
$$\Rightarrow {\text{5}}^{\left(x+3\right)}\text{ = }{\left(5^2\right)}^{(3x-4)}$$
$$\Rightarrow {\text{5}}^{\left(x+3\right)}\text{ = }{{\text{5}}^2}^{(3x-4)}$$
$$\Rightarrow {\text{5}}^{\left(x+3\right)}\text{ = }{\text{5}}^{(6x-8)}$$
$$\Rightarrow x+3=6x-8$$
$$\Rightarrow 5x=11$$
$$\Rightarrow x=\frac{11}{5}$$

$$\frac{2^{n+4}-2\left(2^n\right)}{2\left(2^{n+3}\right)}$$
$$=\frac{2^{n+4}-2^{n+1}}{2^{n+4}}$$
$$=\frac{2^{n+4}}{2^{n+4}}-\frac{2^{n+1}}{2^{n+4}}$$
$$=1-2^{(n+1)-\left(n+4\right)}$$
$$\text{ = 1}-2^{-3}$$
$$\text{ = 1}-\frac{1}{8}$$
$$\text{ = }\frac{7}{8}$$

$$a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}$$
$$=\frac{{\left(\sqrt{5}+1\right)}^2+{\left(\sqrt{5}-1\right)}^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}$$
$$=\frac{2\left[{\left(\sqrt{5}\right)}^2+1\right]}{5-1}$$
$$=\frac{2\left(5+1\right)}{4}$$
$$=3$$
$$a.b=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1$$
$$\text{Put value in expression}$$
$$\frac{a^2+ab+b^2}{a^2-ab+b^2}$$
$$=\frac{{\left(a+b\right)}^2-ab}{{\left(a+b\right)}^2-3ab}$$
$$=\frac{3^2-1}{3^2-3}$$
$$=\frac{9-1}{9-3}$$
$$=\frac{4}{3}$$