Question

Q=>
$log_{3}( {3}^{x} - 8 ) = 2 - x$
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Answer 1

log3(3x−8)=2−x⇒3x−8=32−x⇒3x−8=3x9⇒(3x)2−8.3x−9=0⇒(3x−9)(3x+1)=0

⇒3x=−1 where 3x is always positive.

⇒3x=9⇒x=2.

Answer 2

Answer:

Solution :-

$\large \bold{ log_{3}( {3}^{x} - 8) } = \bold{2 - x} \\ \\ : \to \: {3}^{x} - 8 = {3}^{2 - x} \\ \\ : \to \: \: \: {3}^{x} - 8 = \frac{9}{ {3}^{x} }$

Let ${3}^{x}$ be m

$: \to \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: m - 8 = \frac{9}{m} \\ \\ : \to \: \: \: \: \: \: \: \: \: \: \: \: \: {m}^{2} - 8m = 9 \\ \\ : \to \: \: \: \: \: \: {m}^{2} - 8m - 9 = 0 \\ \\ : \to \: (m - 9)(m + 1) = 0$

(m + 1) is not possible because it gives negative value of x.

$m - 9 = 0 \\ \\ m = 9 \\ \\ {3}^{x} = 9 \\ \\ \bf\implies \:\boxed { \bf \red{x = 2}}$