Question

this is exponets chapter question today is my exam please solve my doubt
12.no question.​

Answer 1

Solutions ⤵️

$12) \: \: \: \:$

Given :- (-2/3)^-3

Find : By what it should be divided to get (4/27)^-2

$solution \:$

(-2/3)^-3/x= (4/27)^-2

(-3/2)^-3/x= (27)^2/4

x= -3/2×-3/2×-3/2/27/4×27/4

x= -1/2×4/27

x= -2/27

$so \: \\ \\ answer \: is \: x = - 2 \div 27 \:$

_________________________________________

$13)$

Given 5^2x+1÷25=125

To find : Value of x

$solution \:$

5^2x+1÷25=125

5^2x+1÷5×5=5×5×5

5^2x+1-2=5^3

Since base of LHS and RHS are equal

therefore

2x-1= 3

2x= 3+1

x= 4/2 => 2

$hence \: \\ \\ \\ x = 2$

Answer 2

Solutions ⤵️

Given :- (-2/3)^-3

Find : By what it should be divided to get (4/27)^-2

solution \:solution

(-2/3)^-3/x= (4/27)^-2

(-3/2)^-3/x= (27)^2/4

x= -3/2×-3/2×-3/2/27/4×27/4

x= -1/2×4/27

x= -2/27

\begin{gathered}so \: \\ \\ answer \: is \: x = - 2 \div 27 \: \end{gathered}

so

answerisx=−2÷27

_________________________________________

Given 5^2x+1÷25=125

To find : Value of x

solution \:solution

5^2x+1÷25=125

5^2x+1÷5×5=5×5×5

5^2x+1-2=5^3

Since base of LHS and RHS are equal

therefore

2x-1= 3

2x= 3+1

x= 4/2 => 2

hence

x=2