Question

hey will u pls help me solving this ..........

what will be there in place of x and how ???​

Answer 1

• Question

$\sf{\bigg(\dfrac{5}{-6}\bigg)^4 ÷ \bigg(\dfrac{5}{-6}\bigg)^x = \bigg[\bigg(\dfrac{5}{-6}\bigg)^2\bigg]^3}$

• Solution

$\sf{\bigg(\dfrac{5}{-6}\bigg)^4 ÷ \bigg(\dfrac{5}{-6}\bigg)^x = \bigg[\bigg(\dfrac{5}{-6}\bigg)^2\bigg]^3}$

$\sf{\bigg(\dfrac{5}{-6}\bigg)^{4 - x} = \bigg[\bigg(\dfrac{5}{-6}\bigg)^2\bigg]^3}$

$\sf{\bigg(\dfrac{5}{-6}\bigg)^{4 - x} = \bigg(\dfrac{5}{-6}\bigg)^{2 \times 3}}$

$\sf{\bigg(\dfrac{5}{-6}\bigg)^{4 - x} = \bigg(\dfrac{5}{-6}\bigg)^{6}}$

Bases are equal, powers are equal.

$\sf{4 - x = 6}$

$\mapsto\sf{ - x = 6 - 4}$

$\mapsto\sf{ - x = 2}$

$\mapsto\sf{x = - 2}$

The value of x = -2

━━━━━━━━━━━

• Know MorE -

$\red{\bigstar}$   Laws of Indices -

• 1st Law (Product Law)

For example -

$:  \implies\sf{ {a}^{2} \times  {a}^{3} =  {a}^{2 + 3}   }$

$:  \implies\sf{ {a}^{5}   }$

• 2nd Law  (Quotient law)

For example -

$: \implies\sf{ \dfrac{a^{2} }{a^{3}} }$

$: \implies\sf{a^{2 - 3} }$

$: \implies\sf{a^{ - 1} }$

• 3rd Law (Power law)

For example -

$: \implies\sf{(a^2)^3}$

$: \implies\sf{(a^{2 \times 3})}$

$: \implies\sf{(a^{6})}$

Answer 2

Given:-

$\sf \bullet {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{4} \div {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{x} = {\bigg\{{\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{2}\bigg \}}^{3}$

Find:-

$\sf \bullet Value \: of \: x$

Solution:-

$\sf :\Longrightarrow {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{4} \div {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{x} = {\bigg\{{\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{2}\bigg \}}^{3}$

$\sf \because {a}^{m} \div {a}^{n} = {a}^{m - n}$

$\sf :\Longrightarrow {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{4 - x}= {\bigg\{{\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{2}\bigg \}}^{3}$

$\sf \because {({a}^{m})}^{n} = {a}^{mn}$

$\sf :\Longrightarrow {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{4 - x}= {\bigg\lgroup \dfrac{5}{ - 6}\bigg \rgroup}^{6}$

$\sf \because {a}^{m}= {a}^{n} \implies m = n$

$\sf :\Longrightarrow 4 - x = 6$

$\sf :\Longrightarrow 4 = 6 + x$

$\sf :\Longrightarrow 4 - 6 =x$

$\sf :\Longrightarrow - 2 =x$

$\underline{\boxed{\sf \therefore The\:value\:of\:x\:is\:-2}}$