Question

Need well explained answer

both answers

Answer 1

Hope it helps
:)
Note:For the 21st sum I have used index to log form conversion hope u know it...

Answer 2

(21.)

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ {a}^{x} = {b}^{y} = {c}^{z}}}$

$\\ \: \: { \huge{.}} \: \: { \bold{ \dfrac{b}{a} = \dfrac{c}{b} }}$

TO PROVE :–

$\\ \: \: { \huge{.}} \: \: { \bold{ \dfrac{2x}{x + z} = \dfrac{y}{x}}}$

SOLUTION :–

• Let –

$\\ \implies { \bold{ {a}^{x} = {b}^{y} = {c}^{z} = \lambda}}$

• We should write this as –

$\\ \implies { \bold{ a = \lambda^{ \frac{1}{x}} \: , \: b= \lambda^{ \frac{1}{y}} \: , \: c= \lambda^{ \frac{1}{z}}}}$

• Now use other condition –

$\\ \implies { \bold{ \dfrac{b}{a} = \dfrac{c}{b} }}$

$\\ \implies { \bold{ {b}^{2} = ac }}$

• Now put the value of a , b and c –

$\\ \implies { \bold{ {(\lambda^{ \frac{1}{y}})}^{2} = (\lambda^{ \frac{1}{x}})(\lambda^{ \frac{1}{z}}) }}$

$\\ \implies { \bold{ {(\lambda^{ \frac{2}{y}})} = (\lambda^{ \frac{1}{x} + \frac{1}{z} })}}$

• Now compare –

$\\ \implies { \bold{ \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}}}$

$\\ \implies { \bold{ \dfrac{2}{y} = \dfrac{x + z}{xz} }}$

• We should write this as –

$\\ \implies { \bold{ \dfrac{2z}{x + z} = \dfrac{y}{x} }}$

$\\ \: \: \: \: \: \: \: { \bold{ \underbrace{ Hence \: \: proved }}}$

$\\ \rule{220}{2}$

(22.)

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ {3}^{x + y} = 81 }}$

$\\ \: \: { \huge{.}} \: \: { \bold{ {(81)}^{x - y} = 3}}$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

• According to the first condition –

$\\ \implies { \bold{ {3}^{x + y} = 81 }}$

• We should write this as –

$\\ \implies { \bold{ {3}^{x + y} = {3}^{4} }}$

• We should write this as –

$\\ \implies { \bold{x + y = 4 \: \: \: \: - - - eq.(1)}}$

• According to the second condition –

$\\ \implies { \bold{ {(81)}^{x - y} = 3}}$

• We should write this as –

$\\ \implies { \bold{ {( {3}^{4} )}^{x - y} = 3}}$

$\\ \implies { \bold{ {(3)}^{4(x - y)} = 3}}$

• Now compare –

$\\ \implies { \bold{x - y = \dfrac{1}{4} \:\:\:\ ---eq.(2)}}$

• Now add eq.(1) and eq.(2) –

$\\ \implies { \bold{x + y + x - y = 4 + \dfrac{1}{4} }}$

$\\ \implies { \bold{2x = \dfrac{17}{4} }}$

$\\ \implies \large { \boxed{ \bold{x = \dfrac{17}{8} }}}$

• Now Using eq.(1) –

$\\ \implies { \bold{ \dfrac{17}{8} + y = 4}}$

$\\ \implies { \bold{ y = 4 - \dfrac{17}{8} }}$

$\\ \implies { \bold{ y = \dfrac{32 - 17}{8} }}$

$\\ \implies \large{ \boxed { \bold{ y = \dfrac{15}{8} } }}$