Question

find the value of n (-4)^n+2×(-4)^7=(-4)^4​

Answer 1

Required Answer:-

Given:

• $\rm {( - 4)}^{n + 2} \times {( - 4)}^{7} = {( - 4)}^{4}$

To find:

• The value of n.

Solution:

We have,

$\rm \implies {( - 4)}^{n + 2} \times {( - 4)}^{7} = {( - 4)}^{4}$

$\rm \implies {( - 4)}^{n + 2 + 7} = {( - 4)}^{4} \: \: \: \: \blue{ \bigg( {x}^{a} \times {x}^{b} = {x}^{a + b} \bigg) }$

$\rm \implies {( - 4)}^{n + 9} = {( - 4)}^{4} \: \: \: \: \blue{ \bigg( {x}^{a} \times {x}^{b} = {x}^{a + b} \bigg) }$

Comparing base, we get,

$\rm \implies n + 9= 4$

$\rm \implies n = 4 - 9$

$\rm \implies n = - 5$

★ Hence, the value of n is -5.

Answer:

• The value of n is -5.

Formulae To Know:

• $\rm {x}^{a} \times {x}^{b} = {x}^{a + b}$
• $\rm ({x}^{a})^{b} = {x}^{ab}$
• $\rm {x}^{0} = 1 \: \: (x \neq 0)$
• $\rm {x}^{y} = \dfrac{1}{ {x}^{ - y} }$
• $\rm \dfrac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b}$
• $\sf \sqrt[y]{x} = {x}^{^{1}/_{y} }$

Answer 2

$\sf{\left(-4\right)^{n+2}\left(-4\right)^7=\left(-4\right)^4}$

Apply Exponent Rule : $\sf{a^b+a^c=a^{b+c}}$

$\to\sf{\left(-4\right)^{n+2+7}=\left(-4\right)^4}$

If $\sf{a^{f\left(x\right)}=a^{g\left(x\right)}}$, then $\sf{f\left(x\right)=g\left(x\right)}$

$\to\sf{n+2+7=4}$

$\to\sf{n+9=4}$

Subtract 9 from Both Sides

$\to\sf{n+9-9=4-9}$

$\to\sf{n=4-9}$

$\to\sf{n=-5}$