Question

$\large{\boxed{\boxed{\sf{Question}}}}$
$\bf \:Solve \: for \: x :- {5}^{ - 4x} \times {5}^{ {x}^{2} + 4} = 1$​

Answer 1

Identity used :-

• $\sf \: {a}^{m} \times {a}^{n} = {a}^{m + n}$
• $\sf \: {a}^{0} = 1$
• $\sf\: {a}^{m} = {a}^{n} ⟹ \: m = n$

$\Large{\underbrace{\sf{\purple{Required\:Solution:}}}}$

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

$\sf:\implies \: {5}^{ - 4x \: + \: {x}^{2} + 4} = {5}^{0}$

$\sf:\implies \: {x}^{2} - 4x + 4 = 0$

$\sf:\implies \: {x}^{2} - 2x - 2x + 4 = 0$

$\sf:\implies \:x(x - 2) - 2(x - 2) = 0$

$\sf:\implies \:(x - 2)(x - 2) = 0$

$\sf:\implies \:x = 2$

_____________________________________________

Answer 2

Step-by-step explanation:

Identity used :-

\begin{gathered}\sf \: {a}^{m} \times {a}^{n} = {a}^{m + n} \\\\\end{gathered}

a

m

×a

n

=a

m+n

\begin{gathered}\sf \: {a}^{0} = 1\\\\\end{gathered}

a

0

=1

\begin{gathered}\sf\: {a}^{m} = {a}^{n} ⟹ \: m = n\\\\\end{gathered}

a

m

=a

n

⟹m=n

\begin{gathered}\Large{\underbrace{\sf{\purple{Required\:Solution:}}}}\\\\\end{gathered}

RequiredSolution:

\begin{gathered}\sf \: {5}^{ - 4x} \times {5}^{ {x}^{2} + 4} = 1\\\\\end{gathered}

5

−4x

×5

x

2

+4

=1

\begin{gathered}\sf:\implies \: {5}^{ - 4x \: + \: {x}^{2} + 4} = {5}^{0}\\\\ \end{gathered}

:⟹5

−4x+x

2

+4

=5

0

\begin{gathered}\sf:\implies \: {x}^{2} - 4x + 4 = 0\\\\\end{gathered}

:⟹x

2

−4x+4=0

\begin{gathered}\sf:\implies \: {x}^{2} - 2x - 2x + 4 = 0\\\\\end{gathered}

:⟹x

2

−2x−2x+4=0

\begin{gathered}\sf:\implies \:x(x - 2) - 2(x - 2) = 0\\\\\end{gathered}

:⟹x(x−2)−2(x−2)=0

\begin{gathered}\sf:\implies \:(x - 2)(x - 2) = 0\\\\\end{gathered}

:⟹(x−2)(x−2)=0

\begin{gathered}\sf:\implies \:x = 2\\\\\end{gathered}

:⟹x=2