Find x- 3x-3-2-3x2-3x-u

3^x 3 =2 √(3^x) Let 3^x= t ⇒ t 3=2 √(t) Squaring both sides, t^2+9 6t= 4t t^2 10t+9=0 t^2 9t t+9=0 t (t 9) 1 (t 9)=0 (t ...

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Byju's Answer

Standard VII

Mathematics

Expression Formation

Find x: 3x...

Question

Find $x$ :
$3^{x} - 3 = 2 \sqrt{3^{x}} (2 \sqrt{3^{x}} = u)$

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Solution

$3^{x} - 3 = 2 \sqrt{3^{x}}$
Let $3^{x} = t$
$\Rightarrow t - 3 = 2 \sqrt{t}$
Squaring both sides,
$t^{2} + 9 - 6 t = 4 t$
$t^{2} - 10 t + 9 = 0$
$t^{2} - 9 t - t + 9 = 0$
$t (t - 9) - 1 (t - 9) = 0$
$(t - 1) (t - 9) = 0$
$t = 1$ or $t = 9$
$3^{x} = 1$ or $3^{x} = 9$
$\Rightarrow x = 0$ or $x = 2$
But $x = 0$ makes $L . H . S < 0$ which is the possible as $R . H . S > 0$
$\Rightarrow x = 2$

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Find x:3x−3=2√3x(2√3x=u)

3x−3=2√3xLet 3x=t⇒t−3=2√tSquaring both sides,t2+9−6t=4tt2−10t+9=0t2−9t−t+9=0t(t−9)−1(t−9)=0(t−1)(t−9)=0t=1 or t=93x=1 or 3x=9⇒x=0 or x=2But x=0 makes L.H.S<0 which is the possible as R.H.S>0⇒x=2

Find $x$ :
$3^{x} - 3 = 2 \sqrt{3^{x}} (2 \sqrt{3^{x}} = u)$