Question

$3^{x-1}$ + $3^{x}$ - 12=0

Answer 1

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

Given equation is

$\rm \: {3}^{x - 1} + {3}^{x} - 12 = 0$

We know,

$\boxed{\rm{  \: {a}^{x - y} = \frac{ {a}^{x} }{ {a}^{y} } \: }}$

So, using this, the above expression can be rewritten as

$\rm \: \dfrac{ {3}^{x} }{3} + {3}^{x} - 12 = 0$

$\rm \: \dfrac{ {3}^{x} + 3. {3}^{x} - 36}{3} = 0$

$\rm \: {3}^{x} + 3. {3}^{x} = 36$

$\rm \: {3}^{x}(1 + 3) = 36$

$\rm \: {3}^{x} \times 4 = 36$

$\rm \: {3}^{x} = 9$

$\rm \: {3}^{x} = {3}^{2}$

$\rm\implies \:x = 2$

Verification

Given equation is

$\rm \: {3}^{x - 1} + {3}^{x} - 12 = 0$

On substituting x = 2, we get

$\rm \: {3}^{2 - 1} + {3}^{2} - 12 = 0$

$\rm \: {3}^{1} + 9 - 12 = 0$

$\rm \: 3 - 3 = 0$

$\rm \: 0 = 0$

Hence, Verified

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

Additional Information :-

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