Question

$25( {5}^{x - 2} ) = {5}^{x}$
PROVE THAT LHS = RHS​

Answer 1

Answer:

$25 = 5 ^{2} \\ 5 ^{2} (5 ^{x - 2} ) = 5 ^{x} \\ bases \: are \: same \\ so \: we \: would \: compare \\ only \: powers \\ 2(x - 2) = x \\ 2x - 4 = x \\ x = 4$

Therefore x=4

Substituting x=4, we get:

${5}^{2} ( {5}^{4 - 2} ) = {5}^{4} \\ 25 \times 25 = 625 \\ 625 = 625$

Therefore, L.H.S.=R.H.S.

Hope it helps

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Answer 2

$\large\underline{\sf{Solution-}}$

Given expression, to prove that

$\rm \: 25 \: [ {5}^{x - 2}] = {5}^{x}$

Consider LHS

$\rm \: 25 \: [ {5}^{x - 2}]$

can be rewritten as

$\rm \: = \: 5 \times 5 \times \: {5}^{x - 2}$

We know,

$\boxed{ \rm{ \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: \: }}$

So, using this law of exponents, we get

$\rm \: = \: {5}^{1 + 1 + x - 2}$

$\rm \: = \: {5}^{2 + x - 2}$

$\rm \: = \: {5}^{x}$

Hence,

$\rm\implies \:\rm \: 25 \: [ {5}^{x - 2}] = {5}^{x}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$