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# Surds & Indices

1.

$$(256)^{0.16}\times (256)^{0.09}=?$$

$$(256)^{0.16}\times (256)^{0.09}=(256)^{(0.16+0.09)}$$

$$=(256)^{0.25}$$

$$=(256)^{(25/100)}$$

$$=(256)^{(1/4)}$$

$$=(4^{4})^{(1/4)}$$

$$=4^{4(1/4)}$$

$$=4^{1}$$

$$=4$$

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2.

$$\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}}=n^{?}$$

$$\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}}$$

$$= \frac{1}{(1+\frac{an}{am})}+\frac{1}{(1+\frac{am}{an})}$$

$$= \frac{am}{(am+an)}+\frac{an}{(am+an)}$$

$$= \frac{(am+an)}{(am+an)}$$

$$= 1$$

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3.

If $$2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}$$, then $$x$$ is equal to:

$$2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}$$

$$\Rightarrow 2^x \times (2^4)^{\frac{2}{5}} = 2^{\frac{1}{5}}$$

$$\Rightarrow 2^x \times 2^{\frac{8}{5}}= 2^{\frac{1}{5}}$$

$$\Rightarrow 2^{(x+ \frac{8}{5})} = 2^{\frac{1}{5}}$$

$$\Rightarrow x + \frac{8}{5} = \frac{1}{5}$$

$$\Rightarrow x = (\frac{1}{5} - \frac{8}{5}) = -\frac{7}{5}$$

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4.

$$(25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5?$$

Let $$(25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5x.$$

Then, $$\frac{(5)^{7.5} \times (5)^{2.5}}{(5^{3})^{1.5}} = 5x$$

$$\Rightarrow \frac{5^{(2 \times 7.5)} \times (5)^{2.5}}{5^{(3 \times 1.5)}} = 5x$$

$$\Rightarrow \frac{5^{15} \times 5^{2.5}}{5^{4.5}} = 5x$$

$$\Rightarrow 5x = 5^{(15+2.5-4.5)}$$

$$\Rightarrow 5x = 5^{13}, \therefore x = 13$$

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5.

$$(25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^?$$

Let $$(25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^{x}.$$

Then, $$\frac{(5^{2})^{7.5}\times(5)^{2.5}}{(5^{3})^{1.5}}=5^{x}$$

$$\Rightarrow \frac{5^{(2\times7.5)}\times5^{2.5}}{5^{(3\times1.5)}}=5^x$$

$$\Rightarrow \frac{5^{15}\times5^{2.5}}{5^{4.5}}=5^x$$

$$\Rightarrow 5^{x}=5^{(15+2.5-4.5)}$$

$$\Rightarrow 5^{x}=5^{13}$$

$$\therefore x=13$$

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6.

If $$m$$ and $$n$$ are whole numbers such that $$m^{n}=121$$, the value of $$(m-1)^{n+1}$$ is:

We know that $$11^{2}=121$$.

Putting m = 11 and n = 2, we get:

$$(m-1)^{n+1}=(11-1)^{(2+1)}=10^{3}=1000$$.

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7.

$$\frac{1}{1+x^{(b-a)}+x^{(c-a)}}+\frac{1}{1+x^{(a-b)}+x^{(c-b)}}+\frac{1}{1+x^{(b-c)}+x^{(a-c)}}=?$$

Given Exp.= $$\frac{1}{(1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}})}+\frac{1}{(1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}})}+\frac{1}{(1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}})}$$

$$=\frac{x^{a}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{b}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{c}}{(x^{a}+x^{b}+x^{c})}$$

$$=\frac{(x^{a}+x^{b}+x^{c})}{(x^{a}+x^{b}+x^{c})}$$

$$=1$$

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8.

$$(0.04)^{-1.5}=?$$

$$(0.04)^{-1.5}=(\frac{4}{100})^{-1.5}$$

$$= (\frac{1}{25})^{-(3/2)}$$

$$= (25)^{(3/2)}$$

$$= (5^{2})^{(3/2)}$$

$$= (5)^{2\times(3/2)}$$

$$= 5^3$$

=125

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9.

If $$\sqrt{(3+3\sqrt{x})} = 2$$, then $$x$$ is equal to:

Exp: On squaring both sides, we get:

$$3+3\sqrt{x} = 4$$ or $$3\sqrt{x} = 1$$

Cubing both sides, we get $$x = (1\times 1\times 1) = 1$$

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10.

$$(25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^?$$

Let $$(25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^x$$

Then, $$\frac{(5^{2})^{7.5}\times (5)^{2.5}}{(5^{3})^{1.5}}=5^x$$

$$\Rightarrow \frac{5^{(2\times7.5)}\times5^{2.5}}{(5^{3\times1.5})}=5^x$$

$$\Rightarrow \frac{5^{15}\times5^{2.5}}{5^{4.5}}=5^x$$

$$\Rightarrow 5^{x}=5^{(15+2.5-4.5)}$$

$$\Rightarrow 5^{x}=5^{13}$$

$$\therefore x=13$$

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