Solve for xi 115-5-115-2- 115x-9ii 292x-1-293- 29x-1

The correct option is A i) 2 ii) 3 (1)(11/5)^ 5 2=(11/5)^x 9 ∴ #10; 7=x 9 ∴ x= 7+9=2 (2)(2/9)^2x+1=(2/9)^ 3=(2/9)^x+1 (2/9)^2x+1 3=( ...

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Byju's Answer

Standard VIII

Mathematics

Finding square of a number using (a+b)^2 identity

Solve for x...

Question

Solve for $x$

(i) ${(\frac{11}{5})}^{- 5} \times {(\frac{11}{5})}^{- 2} = {(\frac{11}{5})}^{x - 9}$
(ii) ${(\frac{2}{9})}^{2 x + 1} \div {(\frac{2}{9})}^{3} = {(\frac{2}{9})}^{x + 1}$

i) 2 i i) 3

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i) - 2 i i) 3

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i) 2 i i) - 3

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i) - 2 i i) - 3

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Solution

The correct option is A $i) 2 i i) 3$
$(1) (\frac{11}{5})^{- 5 - 2} = (\frac{11}{5})^{x - 9} ∴ - 7 = x - 9 ∴ x = - 7 + 9 = 2 (2) (\frac{2}{9})^{2 x + 1} = (\frac{2}{9})^{- 3} = (\frac{2}{9})^{x + 1} (\frac{2}{9})^{2 x + 1 - 3} = (\frac{2}{9})^{x + 1} 2 x - 2 = x + 1 ∴ x = 3$

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Similar questions

Q. Find the values of

x .

(i)

3^{x} \times 3^{- 2} = 3^{2}

(ii)

5^{- 3} \div 5^{x} = 5^{2}

(iii)

{(\frac{11}{5})}^{- 5} \times {(\frac{11}{5})}^{- 2} = {(\frac{11}{5})}^{x - 9}

Solve (i) (3x + 24) / (2x + 7) = 2: (ii) (1 − 9y) / (11 − 3y) = (5/8).

Add (i) $(2 \sqrt{3} - 5 \sqrt{2}) a n d (\sqrt{3} + 2 \sqrt{2})$

(ii) $(2 \sqrt{2} + 5 \sqrt{3} - 7 \sqrt{5}) a n d (3 \sqrt{3} - \sqrt{2} + \sqrt{5})$

(iii) $(\frac{2}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + 6 \sqrt{11}) a n d (\frac{1}{3} \sqrt{7} + \frac{3}{2} \sqrt{2} - \sqrt{11})$

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Solve :

(i)

4 \times \frac{1}{3}

(ii)

2 \times \frac{6}{7}

(iii)

11 \times \frac{4}{7}

(iv)

20 \times \frac{4}{5}

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Solve for x(i) (115)−5×(115)−2=(115)x−9(ii) (29)2x+1÷(29)3=(29)x+1

The correct option is A i)2 ii)3(1)(115)−5−2=(115)x−9∴−7=x−9∴x=−7+9=2(2)(29)2x+1=(29)−3=(29)x+1(29)2x+1−3=(29)x+12x−2=x+1∴x=3

Solve for $x$

(i) ${(\frac{11}{5})}^{- 5} \times {(\frac{11}{5})}^{- 2} = {(\frac{11}{5})}^{x - 9}$
(ii) ${(\frac{2}{9})}^{2 x + 1} \div {(\frac{2}{9})}^{3} = {(\frac{2}{9})}^{x + 1}$

The correct option is A $i) 2 i i) 3$
$(1) (\frac{11}{5})^{- 5 - 2} = (\frac{11}{5})^{x - 9} ∴ - 7 = x - 9 ∴ x = - 7 + 9 = 2 (2) (\frac{2}{9})^{2 x + 1} = (\frac{2}{9})^{- 3} = (\frac{2}{9})^{x + 1} (\frac{2}{9})^{2 x + 1 - 3} = (\frac{2}{9})^{x + 1} 2 x - 2 = x + 1 ∴ x = 3$