Question

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Solve this attachment question
NOTE >Solve both questions 18 and 19.
Right answer will mark as brainlist ✌️

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Answer 1

18.

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

Value of (b) given=a^x

Then

${(b)}^{y} = c\\ \\ b = {a}^{x} \\ \\ ( {a}^{x} ) {}^{y} = c \\ \\ {a}^{xy} = c$

So

C=a^xy

Now Given

${(c)}^{z} = a$

Putting Value of C as a^xy

$( {a}^{xy}) {}^{z} = a \\ \\ {a}^{xyz} = a$

Remember When Base is Same in LHS and RHS of Equation Then Powers are Written as Equation.

${a}^{xyz} = a \times 1 \\ \\ xyz = 1$

Proved√√

19.

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

Comparing Both Sides

${2}^{x + 1} = {2}^{5} \\ \\ x + 1 = 5 \\ \\ x = 5 - 1 \\ \\ x = 4$

${3}^{y - 1} = {3}^{5} \\ \\ y - 1 = {5}^{} \\ \\ y = 5+ 1 \\ \\ y = 6$

X=4

Y=6

3x-2y

3×4-2×6

12-12

=0

Answer=0

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

18 ]

GIVEN :

$\sf{{a}^{x} = b \: \: \: ,{b}^{y} = c \: \: \: ,{c}^{z} = a}$

TO PROVE :

$\implies \sf{xyz = 1}$

PROOF :

$\implies \: \sf{ {a}^{x} = b \: \: \rightarrow \: (1)}$

$\implies \: \sf{ {b}^{y} = c} \: \:$

Putting the value of 'b' from equation (1) we get,

$\implies \: \sf{ { ({a}^{x}) }^{y} = c} \: \:$

$\implies \: \sf{ { {a}^{xy} } = c} \: \: \: \rightarrow \: (2)$

$\sf {\implies \: {c}^{z} = a}$

Putting the value of 'c' from equation (2) we get,

$\sf {\implies \: {( {a}^{xy} )}^{z} = a}$

$\sf {\implies \: { {a}^{xyz} }= a}$

$\implies \: \sf{xyz \: = \: 1}$

$\boxed{ \sf{Hence, proved.}}$

19]

GIVEN :

$\sf{ {2}^{ x + 1} . \: {3}^{y - 1} = {(2 \: . \: 3)}^{5} }$

TO FIND :

$= \: \sf{3x - 2y}$

SOLUTION :

$\implies \: \sf{ {2}^{ x + 1} . \: {3}^{y - 1} = {(2 \: . \: 3)}^{5}}$

$\implies \: \sf{ {2}^{ x + 1} . \: {3}^{y - 1} = { {2}^{5} . \: {3}^{5}}}$

On comparing both sides we get,

$\sf{ \implies \: {2}^{x + 1} = {2}^{5} \: \: \rightarrow(1) \: and \: \: {3}^{y - 1} = {3}^{5} \rightarrow(2)}$

Solving (1) we get,

$\sf{ \implies \: {2}^{x + 1} = {2}^{5}}$

$\sf{ \implies \: {x + 1} = {5}}$

$\sf{ \implies \: {x } = 4}$

Solving (2) we get,

$\sf{ \implies \: {3}^{y - 1} = {3}^{5} }$

$\sf{ \implies \: y - 1 = 5}$

$\sf{ \implies \: y = 6}$

Now, putting the values of 'x' and 'y' in our required we get,

$= \: \sf{3x - 2y}$

$= \: \sf{3(4)- 2(6)}$

$= 12 - 12$

$= 0$

$\boxed { \sf{ \:Hence, \: 3x - 2y} = 0}$