How do you solve $4 [9^{x - 1}] = 108$ $4[9^(x-1)] = 108$ ?

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Answer 1 · 2018-04-18T05:06:02

$x = \frac{5}{2}$ $x=5/2$

Here,

$4 [9^{x - 1}] = 108$ $4[9^(x-1)]=108$

Dividing both sides by $4$ $4$

$\frac{4 [9^{x - 1}]}{4} = \frac{108}{4}$ $(4[9^(x-1)])/4=108/4$

$\Rightarrow [9^{x - 1}] = 27$ $=>[9^(x-1)]=27$

$\Rightarrow {(3^{2})}^{x - 1} = 3^{3}$ $=>(3^2)^(x-1)=3^3$

$\Rightarrow 3^{2 x - 2} = 3^{3}$ $=>3^(2x-2)=3^3$

$\Rightarrow 2 x - 2 = 3$ $=>2x-2=3$

$\Rightarrow 2 x = 5$ $=>2x=5$

$\Rightarrow x = \frac{5}{2}$ $=>x=5/2$

How do you solve 4[9x−1]=1084[9^(x-1)] = 108?