Maths M.C.Q

We have to find the value of $(256)^{0.16} \times(256)^{0.09}$. So,
By using law of rational exponents
$a^m \times a^n=a^{m+n} \text { we get }$
$(265)^{0.16} \times(256)^{0.09}=(256)^{0.16} \times(256)^{0.09}$
$=(256)^{0.16+0.09}$
$=256^{0.25}$
$=256^{\frac{25}{100}}$
$(256)^{0.16} \times(256)^{0.09}=2^{8 \times \frac{25}{100}}$
$=2^{8 \times \frac{25}{100}}$
$=2^{8 \times \frac{1}{4}}$
$=2^{8 \times \frac{1}{4}}=4$
The value of $(256)^{0.16} \times(256)^{0.09}$ is $4$
Hence the correct choice is $a$.

Given $\text{x}-\text{y}^{\text{x}-\text{y}}$
Here $\text{x}=\text{2},\ \text{y}=-2$
By substituting in $\text{x}-\text{y}^{\text{x}-\text{y}}$ we get
$\text{x}-\text{y}^{\text{x}-\text{y}}=2-(-2)^{2-(-2)}$
$=2-(-2)^{2+2}$
$=2-(-2)^4$
$=2-(16)$
$=-14$
The value of $\text{x}-\text{y}^{\text{x}-\text{y}}$ is $-14$
Hence the correct choice is $d$.

Find the value of $\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}^{\text{mn}}.$
So,
$\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}^{\text{mn}}=\bigg\{\sqrt[\text{m}]{\text{a}^{\frac{1}{\text{n}}}}\bigg\}^{\text{mn}}$
$=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\bigg\}^{\text{mn}}$
$=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\times\text{m}\times\text{n}\bigg\}$
$\Rightarrow\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\times\text{m}\times\text{n}\bigg\}$
$\Rightarrow\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}=\text{a}$
Hence the correct choice is $b$.

We have to find the value of $x$
Given $9^{x+2}=240+9^x$
$9^x \times 9^2=240+9^x$
$9^2=\frac{240}{9^{\text{x}}}+\frac{9^{\text{x}}}{9^{\text{x}}}$
$81=\frac{240}{9^{\text{x}}}+1$
$81-1=\frac{240}{9^\text{x}}$
$80=\frac{240}{9^\text{x}}$
$9^\text{x}\times80=240$
$9^\text{x}=\frac{240}{80}$
$3^{2\text{x}}=3$
$3^{2\text{x}}=3^1$
By equating the exponents we get
$2​​\text{x}=1$
$\text{x}=\frac{1}{2}$
$\text{x}=0.5$
Hence the correct alternative is $a$.

We have to find $x^3$ provided $x^2$ $= 2$ So,
By raising both sides to the power $\frac{1}{2}$
$\text{x}^{2\times\frac{1}{2}}=2^{\frac{1}{2}}$
$\text{x}^{2\times\frac{1}{2}}=\sqrt{2}$
$\text{x}=\sqrt{2}$
By substituting $\text{x}=\sqrt{2}$ in $x^3$ we get
$\text{x}^3\big(\sqrt{2}\big)^3$
$=\sqrt{2}\times\sqrt{2}\times\sqrt{2}$
$=2\sqrt{2}$
The value of $x^3$ is $2\sqrt{2}$
Hence the correct choice is $b$.

We have to find the value of $10^{-y}$ Given that $10^{2 y}=25$, therefore,
$10^{2 y}=25$
$\left(10^y\right)^2=5^2$
$\left(10^y\right)^{2 \times \frac{1}{2}}=5^{2 \times \frac{1}{2}}$
$\frac{10^y}{1}=\frac{5}{1}$
$\frac{1}{5}=\frac{1}{10^y}$
$\frac{1}{5}=10^y$
Hence the correct option is $d$.

We have to find the value of $3^x$ provided $\left(2^3\right)^2=4^x$
So,
$2^{3 \times 2}=2^{2 x}$
$2^6=2^{2 x}$
By equating the exponents we get
$6=2 x$
$\frac{6}{2}=x$
$3=x$
By substituting in $3^{\mathrm{x}}$ we get
$3^x=3^3$
$=27$
The value of $3^{\mathrm{x}}$ is $27$
Hence the correct choice is $d$.

We have to simplify $\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^\text{n}-2\times5^{\text{n}+1}}$
Taking $5^n$ as a common factor we get
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^\text{n}-2\times5^{\text{n}+1}}=\frac{5^​​​​\text{n}(5^2-6\times5^1)}{5^\text{n}(13-2\times5^1)}$
$=\frac{5^\text{n}(25-30)}{5^\text{n}(13-10)}$
$=\frac{-5}{3}$
Hence the correct alternative is $b$.