Answer:
[tex]x=1[/tex]
Step-by-step explanation:
We are solving for x in the exponential equation:
[tex]100^x = \left(\!\dfrac{1}{10}\!\right)^{\!(x-3)}[/tex]
First, we can rewrite both exponent bases in terms of 10:
- [tex]100 = 10^{2}[/tex]
- [tex]1/10 = 10^{-1}[/tex]
↓↓↓
[tex](10^2)^x = (10^{-1})^{\!(x-3)}[/tex]
Next, we can apply exponent rules to simplify both sides:
[tex]10^{2x} = 10^{-(x-3)}[/tex]
[tex]10^{2x} = 10^{(3-x)}[/tex]
Now that both sides have a base of 10, we can take the common log of both sides:
[tex]\log(10^{2x}) = \log\!\left(10^{(3-x)}\right)[/tex]
[tex]2x = 3-x[/tex]
The equation now just requires algebraic manipulation to solve for x:
[tex]3x=3[/tex]
[tex]\boxed{x=1}[/tex]
