Question 99(iii)

Find the value of x, so that:
$(2^{- 1} + 4^{- 1} + 6^{- 1} + 8^{- 1})^{x} = 1$

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Solution

we have $(2^{- 1} + 4^{- 1} + 6^{- 1} + 8^{- 1})^{x} = 1$
Using law of exponents $a^{- m} = \frac{1}{a^{m}}$
Then ${(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8})}^{x} = 1 \Rightarrow {(\frac{12 + 6 + 4 + 3}{24})}^{x} = 1 \Rightarrow {(\frac{25}{24})}^{x} = 1$
this can be possible only if $x = 0$ , Since $a^{0} = 1$

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Question 99(iii)

Find the value of x, so that:
$(2^{- 1} + 4^{- 1} + 6^{- 1} + 8^{- 1})^{x} = 1$

Prove that: $t a n^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} c o s^{- 1} x; - \frac{1}{\sqrt{2}} \leq x \leq 1$ .

OR If $t a n^{- 1} (\frac{x - 2}{x - 4}) + t a n^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}$ , find the value of x.

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Degree of a Polynomial

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Question 99(iii) Find the value of x, so that: (2−1+4−1+6−1+8−1)x=1

we have (2−1+4−1+6−1+8−1)x=1 Using law of exponents a−m=1am Then (12+14+16+18)x=1⇒(12+6+4+324)x=1⇒(2524)x=1 this can be possible only if x=0, Sincea0=1

Question 99(iii)

Find the value of x, so that:
$(2^{- 1} + 4^{- 1} + 6^{- 1} + 8^{- 1})^{x} = 1$