Answer:
[tex]x = 1 - log_{2}(5) [/tex]
Step-by-step explanation:
[tex] {2}^{x + 2} = 9 ( {2}^{x} ) - 2[/tex]
Using the rules of indices
That's
[tex] {x}^{a + b} = {x}^{a} \times {x}^{b} [/tex]
So we have
[tex] {2}^{x + 2} = {2}^{x} \times {2}^{2} = 4( {2}^{x} )[/tex]
So we have
[tex]4( {2}^{x}) = 9( {2}^{x} ) - 2[/tex]
Let
[tex] {2}^{x} = y[/tex]
We have
4y = 9y - 2
4y - 9y = - 2
- 5y = - 2
Divide both sides by - 5
[tex]y = \frac{2}{5} [/tex]
But
[tex] {2}^{x} = \frac{2}{5} [/tex]
Take logarithm to base 2 to both sides
That's
[tex] log_{2}( {2}^{x} ) = log_{2}( \frac{2}{5} ) [/tex]
[tex] log_{2}(2) ^{x} = x log_{2}(2) [/tex]
[tex] log_{2}(2) = 1[/tex]
So we have
[tex]x = log_{2}( \frac{2}{5} ) [/tex]
Using the rules of logarithms
That's
[tex] log( \frac{x}{y} ) = log(x) - log(y) [/tex]
Rewrite the expression
That's
[tex]x = log_{2}(2) - log_{2}(5) [/tex]
But
[tex] log_{2}(2) = 1[/tex]
So we have the final answer as
[tex]x = 1 - log_{2}(5) [/tex]
Hope this helps you
