Question

please answer this question ​

Answer 1

Question :

Find the value of x.

${( \frac{7}{4} )}^{ - 3} \times { (\frac{7}{4} )}^{ - 5} = { (\frac{7}{4} )}^{x - 2}$

Solution :

${( \frac{7}{4} )}^{ - 3} \times { (\frac{7}{4} )}^{ - 5} = { (\frac{7}{4} )}^{x - 2}$

$= > { (\frac{7}{4} )}^{ - 3 - 5} = { (\frac{7}{4} )}^{x - 2} \\ = > - 3 - 5 = x - 2 \\ = > - 8 = x - 2 \\ = > - x = - 2 + 8 \\ = > - x = 6 \\ = > x = - 6$

Therefore ,the value of x is -6.

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More information :-

Some identities of Indices :

• ${a}^{m} . {a}^{n} = {a}^{m + n}$
• $\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$
• $( {a}^{m})^{n} = {a}^{mn}$
• ${( \frac{a}{b}) }^{m} = \frac{ {a}^{m} }{ {b}^{m} }$

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Answer 2

Question :

Find the value of x.

$\tt \longmapsto {(\dfrac{7}{4})}^{ - 3} \times { (\dfrac{7}{4}) }^{ - 5} = {(\dfrac{7}{4})}^{ x - 2}$

AnsWer :

x = -6.

Solution :

We have, Equation.

$\tt \longmapsto {(\dfrac{7}{4})}^{ - 3} \times { (\dfrac{7}{4}) }^{ - 5} = {(\dfrac{7}{4})}^{ x - 2}$

$\tt\longmapsto {m}^{a} \times {m}^{b} = {m}^{a + b}$

Note : Base value of LHS and RHS same then power be added.

$\rule{90}1$

$\tt \longmapsto - 3 + ( - 5) = x - 2.$

$\tt \longmapsto - 3 - 5 = x - 2.$

$\tt \longmapsto - 8= x - 2.$

$\tt \longmapsto - 6= x .$

$\tt \longmapsto x = - 6.$

Therefore, the value of be -6.

Some Properties.

• a^x * a^y = a^x+y.
• a^x/a^y = a^x-y.
• (a^x)^y = ( a )^xy => a^xy