(vii) log3 (3* - 8) = 2-x
Answers
Answer:
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
\begin{gathered}{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} }\end{gathered}
3
x
−8=3
2−x
3
x
−8=3
2
×3
−x
3
x
−8=9×(
3
1
)
x
3
x
−8=9×
3
x
1
Let
{3}^{x} = a3
x
=a
Then
\begin{gathered}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)\end{gathered}
a−8=9×
a
1
a−8=
a
9
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}}}}
x=2
Step-by-step explanation:
(( ,.,,,,,,..it work...,,,,..))
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