Question

`{2}^(x + 2) = 9 * ( {2}^(x) ) - 2`
How is this solved pls
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Answer 1

`x = 1 - log_(2)(5)`

Explanation:

`{2}^(x + 2) = 9 ( {2}^(x) ) - 2`

Using the rules of indices

That's

`{x}^(a + b) = {x}^(a) * {x}^(b)`

So we have

`{2}^(x + 2) = {2}^(x) * {2}^(2) = 4( {2}^(x) )`

So we have

`4( {2}^(x)) = 9( {2}^(x) ) - 2`

Let

`{2}^(x) = y`

We have

4y = 9y - 2

4y - 9y = - 2

- 5y = - 2

Divide both sides by - 5

`y = (2)/(5)`

But

`{2}^(x) = (2)/(5)`

Take logarithm to base 2 to both sides

That's

`log_(2)( {2}^(x) ) = log_(2)( (2)/(5) )`

`log_(2)(2) ^(x) = x log_(2)(2)`

`log_(2)(2) = 1`

So we have

`x = log_(2)( (2)/(5) )`

Using the rules of logarithms

That's

`log( (x)/(y) ) = log(x) - log(y)`

Rewrite the expression

That's

`x = log_(2)(2) - log_(2)(5)`

But

`log_(2)(2) = 1`

So we have the final answer as

`x = 1 - log_(2)(5)`

Hope this helps you