Find x- if i- 1-4 -4- 1-4 -8- 1-4 -4x-ii- -1-2 -19- -1-2 8- -1-2 -2x-1-iii- 3-2 -3- 3-2 5- 3-2 2x-1-iv 2-5 -3- 2-5 15- 2-5 2-3x-v- 5-4 -x- 5-4 -4- 5-4 5-vi- 8-3 2x-1- 8-3 5- 8-3 x-2

(i) ( 1/4 )^ 4× ( 1/4 )^ 8= ( 1/4 )^ 4x ⇒ ( 1/4 )^ 4 8= ( 1/4 )^ 4x ⇒ ( 1/4 )^ 12= ( 1/4 )^ 4x Comparing we get: 4x= 12⇒ x= 12/ 4=3 ∴ x=3 (ii) ( 1/2 )^ ...

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Byju's Answer

Standard VIII

Mathematics

Exponents with like Bases

Find x, if i ...

Question

Find x, if

$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

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Solution

$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} \Rightarrow {(\frac{1}{4})}^{- 4 - 8} = {(\frac{1}{4})}^{- 4 x} \Rightarrow {(\frac{1}{4})}^{- 12} = {(\frac{1}{4})}^{- 4 x} C o m p a r i n g w e g e t : - 4 x = - 12 \Rightarrow x = \frac{- 12}{- 4} = 3 ∴ x = 3 (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} \Rightarrow {(\frac{- 1}{2})}^{- 19 - 8} = {(\frac{- 1}{2})}^{- 2 x + 1} \Rightarrow {(\frac{- 1}{2})}^{- 27} = {(\frac{- 1}{2})}^{- 2 x + 1} {(\frac{3}{2})}^{2} = {(\frac{3}{2})}^{2 x + 1} C o m p a r i n g, w e g e t : - 2 x + 1 = - 27 - 2 x = - 27 - 1 = - 28 x = \frac{- 28}{- 2} = 14 ∴ x = 14 (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} \Rightarrow {(\frac{3}{2})}^{- 3 + 5} = {(\frac{3}{2})}^{2 x + 1} \Rightarrow 2 x = 2 - 1 = 1 \Rightarrow x = \frac{1}{2} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} \Rightarrow {(\frac{2}{5})}^{- 3 + 15} = {(\frac{2}{5})}^{2 + 3 x} \Rightarrow {(\frac{2}{5})}^{12} = {(\frac{2}{5})}^{2 + 3 x} C o m p a r i n g, w e g e t : 2 + 3 x = 12 3 x - = 12 - 2 = 10 x = \frac{10}{3} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} \Rightarrow {(\frac{5}{4})}^{- x + 4} = {(\frac{5}{4})}^{5} C o m p a r i n g, w e g e t : - x + 4 = 5 \Rightarrow= 5 - 4 = 1 ∴ x = - 1 (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2} \Rightarrow {(\frac{8}{3})}^{2 x + 1 + 5} = {(\frac{8}{3})}^{x + 2} C o m p a r i n g, w e g e t : 2 x + 1 + 5 = x + 2 \Rightarrow 2 x - x = 2 - 1 - 5 \Rightarrow x = - 4 ∴ x = - 4$

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Similar questions

Q. Find x, if
(i)

{(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x}

(ii)

{(\frac{- 1}{2})}^{- 19} \times {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1}

(iii)

{(\frac{3}{2})}^{- 3} \times {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1}

(iv)

{(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x}

(v)

{(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5}

(vi)

{(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}

Solve :

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