Question

Solve the following question

$1 + {3}^{ \frac{x}{2} } = {2}^{x}$
​

Answer 1

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

Given equation is

$\rm :\longmapsto\:1 + {\bigg[3\bigg]}^{\dfrac{x}{2} } = {2}^{x}$

can be further rewritten as

$\rm :\longmapsto\: {1}^{x} + {\bigg[3\bigg]}^{\dfrac{x}{2} } = {2}^{x}$

can be further rewritten as

$\rm :\longmapsto\: \dfrac{{1}^{x} + {\bigg[3\bigg]}^{\dfrac{x}{2} }}{ {2}^{x} } = 1$

$\rm :\longmapsto\:\dfrac{ {1}^{x} }{ {2}^{x} } + \dfrac{{\bigg[3\bigg]}^{\dfrac{x}{2} }}{ {2}^{x} } = 1$

$\rm :\longmapsto\: {\bigg[\dfrac{1}{2} \bigg]}^{x} + \dfrac{ {( \sqrt{3})}^{x} }{ {2}^{x} } = 1$

$\rm :\longmapsto\: {\bigg[\dfrac{1}{2} \bigg]}^{x} + {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{x} = 1$

We know, from Trigonometric ratios of Standard angles,

$\purple{\rm :\longmapsto\:sin\bigg[\dfrac{\pi}{3} \bigg] = \dfrac{ \sqrt{3} }{2}}$

and

$\purple{\rm :\longmapsto\:cos\bigg[\dfrac{\pi}{3} \bigg] = \dfrac{1}{2}}$

So, using this ,we get

$\rm :\longmapsto\: {\bigg(cos\dfrac{\pi}{3} \bigg) }^{x} + {\bigg(sin\dfrac{\pi}{3} \bigg) }^{x} = 1$

We know,

$\purple{\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x = 1 \: \: \forall \: x \: \in \: R}$

$\bf\implies \:x \: = \: 2$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Question:-

Solve the equation:

$\sf \large 1 + 3 {}^{ \frac{x}{2} } = 2 {}^{x} .$

Given:-

The Equation

$\sf \large 1 + 3 {}^{ \frac{x}{2} } = 2 {}^{x} .$

Solution:-

We have,

$\sf \large 1 + 3 {}^{ \frac{x}{2} } = 2 {}^{x}$

Dividing the both sides by 2^x, then we obtain.

$\sf \large ⇒ \frac{1}{2 {}^{x} } + \frac{3 {}^{ \frac{3}{2} } }{2 {}^{x} } = 1$

$\sf \large ⇒ \bigg( \frac{1}{2} \bigg) {}^{x} + \bigg( \frac{3}{4} \bigg) {}^{ \frac{x}{2} } = 1$

$\sf \large ⇒\bigg( \frac{1}{2} \bigg ) {}^{x} + \bigg( \frac{ \sqrt{3} }{2} \bigg) {}^{x} = 1$

$\sf \large⇒(cos \frac{\pi}{3} ) {}^{x} + (sin \frac{\pi}{3} ) {}^{x} = 1$

which us possible only when x = 2.

Answer:-

Hence, x = 2 is only solution of this equation.

Hope you have satisfied. ⚘