Question

solve: log3(3^x - 8) =2-x​

Answer 1

Step-by-step explanation:

$log_{3}( {3}^{x} - 8 ) = 2 - x$

As We Know

.If

log to base the _b(c)=a

Then c=b^a

Then,Taking Base 3 into RHS

${3}^{x} - 8 = {3}^{2 - x} \\ \\ {3}^{x} - 8 = 3 {}^{2} \times {3}^{ - x} \\ \\ {3}^{x} - 8 = 9 \times ( \frac{1}{3} ) ^{x} \\ \\ {3}^{x} - 8 = 9 \times \frac{1}{3 {}^{x} }$

Let

${3}^{x} = a$

Then

$a - 8 = 9 \times \frac{1}{a} \\ \\ a - 8 = \frac{9}{a} \\ \\ a(a - 8) = 9 \\ \\ {a}^{2} - 8a - 9 = 0 \\ \\ {a}^{2} - 9a + a - 9 = 0 \\ \\ a(a - 9) + 1(a - 9) \\ \\ (a + 1)(a - 9)$

Now

a+1=0

a=-1

Substituting 3^x on place of a

3^x=-1

We will not anything with these Equation.

Now

a-9=0

a=9

Substituting value of a=3^x

3^x=9

3^x=3²

X=2

So ,X=2

$\boxed{ \mathbf{ \huge{ \red{x = 2}}}}$