Question

answer please in solution please not options ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Pair of equations are

$\rm :\longmapsto\: {3}^{x + y} = 81 - - - (1)$

and

$\rm :\longmapsto\: {81}^{x - y} = 3 - - - (2)$

Consider equation (1),

$\rm :\longmapsto\: {3}^{x + y} = 81$

$\rm :\longmapsto\: {3}^{x + y} = {3}^{4}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \because\: {a}^{x} = {a}^{y} \: \implies \: x = y}}$

$\rm :\longmapsto\:x + y = 4$

$\rm :\implies\:x = 4 - y - - - (3)$

Now,

Consider equation (2), we get

$\rm :\longmapsto\: {81}^{x - y} = 3$

$\rm :\longmapsto\: { {(3}^{4} )}^{x - y} = 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \because\: { {(x}^{m} )}^{n} = {x}^{mn}}}$

$\rm :\longmapsto\: {3}^{4(x - y)} = {3}^{1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \because\: {a}^{x} = {a}^{y} \: \implies \: x = y}}$

$\rm :\longmapsto\:4(x - y) = 1$

$\rm :\longmapsto\:4x - 4y = 1$

$\rm :\longmapsto\:4(4 - y) - 4y = 1 \: \: \{using \: equation \: (1) \}$

$\rm :\longmapsto\:16 - 4y- 4y = 1$

$\rm :\longmapsto\:16 - 8y = 1$

$\rm :\longmapsto\: - 8y = 1 - 16$

$\rm :\longmapsto\: - 8y = - 15$

$\bf\implies \:y = \dfrac{15}{8} = 1\dfrac{7}{8}$

On substituting the value of y in equation (1), we get

$\rm :\longmapsto\:x + \dfrac{15}{8} = 4$

$\rm :\longmapsto\:x = 4 - \dfrac{15}{8}$

$\rm :\longmapsto\:x = \dfrac{32 - 15}{8}$

$\rm :\longmapsto\:x = \dfrac{17}{8}$

$\bf\implies \:x = 2\dfrac{1}{8}$

Hence,

• Option (4) is correct.

Answer 2

Step-by-step explanation:

you are a student of Allen advance batch

i am also not able to do this i am also in advance batch