The value of $x$ , when $(256)^{- (4 - x)} = 2^{- 20}$ is :

\frac{7}{2}

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\frac{5}{2}

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\frac{3}{2}

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\frac{1}{2}

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Solution

The correct option is C $\frac{3}{2}$
$(256)^{- (4 - x)} = 2^{- 20}$
$\Rightarrow (16^{2})^{- (4 - x)} = 2^{- 20}$
$\Rightarrow (16)^{- (8 - 2 x)} = 2^{- 20}$
$\Rightarrow (2^{4})^{- (8 - 2 x)} = 2^{- 20}$
$\Rightarrow (2)^{- (32 - 8 x)} = 2^{- 20}$
$\Rightarrow 32 - 8 x = 20$
$\Rightarrow 12 = 8 x$
$\Rightarrow x = 3 / 2$

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Similar questions

Q. Question 7
Find the value of

\frac{4}{(216)^{- \frac{2}{3}}} + \frac{1}{(256)^{- \frac{3}{4}}} + \frac{2}{(243)^{- \frac{1}{5}}}

Solve the following equations for x:

(i) $7^{2 x + 3} = 1$

(ii) $2^{x + 1} = 4^{x - 3}$

(iii) $2^{5 x + 3} = 8^{x + 3}$

(iv) $4^{2 x} = \frac{1}{32}$

(v) $4^{x - 1} \times (0.5)^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $2^{3 x - 7} = 256$

Q. Evaluate the following:
(i)

2 x^{3} + 2 x^{2} - 7 x + 72, when x = \frac{3 - 5 i}{2}

(ii)

x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i

(iii)

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, when x = - 1 + i \sqrt{2}

(iv)

x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}}

(v)

2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, when x = - 2 - \sqrt{3} i

Evaluate the following :

$(i) 2 x^{3} + 2 x^{2} - 7 x + 72, w h e n x = \frac{3 - 5 i}{2} (i i) x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, w h e n x = 3 + 2 i (i i i) x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, w h e n x = - 1 + i \sqrt{2} (i v) x^{6} + x^{4} + x^{2} + 1, w h e n x = \frac{1 + i}{\sqrt{2}} (v) 2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, w h e n x = - 2 - \sqrt{3} i$

Statement 1: The value of the polynomial ${x}^{5}+ 2{x}^{4}+ 3{x}^{3}+ {x}^{2}– 7x + 8$ at x = -1 is 14.

Statement 2: The polynomial ${x}^{5}+ 2{x}^{4}+ 3{x}^{3}+ {x}^{2}– 7x + 8$ when divided by x + 1, gives 14 as remainder.

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The value of x , when (256)−(4−x)=2−20 is :

The correct option is C 32(256)−(4−x)=2−20⇒(162)−(4−x)=2−20⇒(16)−(8−2x)=2−20⇒(24)−(8−2x)=2−20⇒(2)−(32−8x)=2−20⇒32−8x=20⇒12=8x⇒x=3/2

The value of $x$ , when $(256)^{- (4 - x)} = 2^{- 20}$ is :