To solve the equation [tex]\(2^{-x} = 10^{1+x}\)[/tex], let's follow these steps:
1. Rewrite the Equation Using Logarithms:
Our goal is to isolate [tex]\(x\)[/tex]. To do this, we can take the logarithm of both sides of the equation. Let's use the common logarithm ([tex]\(\log_{10}\)[/tex]) for simplicity:
[tex]\[
\log_{10}(2^{-x}) = \log_{10}(10^{1+x})
\][/tex]
2. Apply Logarithm Properties:
Use the properties of logarithms to simplify both sides of the equation. Recall the properties [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex]:
[tex]\[
-x \cdot \log_{10}(2) = (1 + x) \cdot \log_{10}(10)
\][/tex]
3. Simplify Further:
Since the logarithm of 10 in base 10 ([tex]\(\log_{10}(10)\)[/tex]) is 1, this simplifies our equation:
[tex]\[
-x \cdot \log_{10}(2) = 1 + x
\][/tex]
4. Isolate [tex]\(x\)[/tex]:
To isolate [tex]\(x\)[/tex], let's move all terms involving [tex]\(x\)[/tex] to one side of the equation:
[tex]\[
-x \cdot \log_{10}(2) - x = 1
\][/tex]
Factor out [tex]\(x\)[/tex] from the left side:
[tex]\[
x \cdot (-\log_{10}(2) - 1) = 1
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
Isolate [tex]\(x\)[/tex] by dividing both sides of the equation by [tex]\((- \log_{10}(2) - 1)\)[/tex]:
[tex]\[
x = \frac{1}{-\log_{10}(2) - 1}
\][/tex]
6. Simplify the Expression:
Notice that [tex]\(\log_{10}(20)\)[/tex] can be rewritten using logarithm properties:
[tex]\[
\log_{10}(20) = \log_{10}(2 \cdot 10) = \log_{10}(2) + \log_{10}(10) = \log_{10}(2) + 1
\][/tex]
Therefore:
[tex]\[
-\log_{10}(2) - 1 = -(\log_{10}(2) + 1) = -\log_{10}(20)
\][/tex]
Substitute this back into our expression for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{1}{-\log_{10}(20)} = -\frac{1}{\log_{10}(20)}
\][/tex]
7. Rewrite the Final Answer:
Finally, using [tex]\(\log_{10}(10) = 1\)[/tex], our final solution simplifies to:
[tex]\[
x = \frac{-\log(10)}{\log(20)}
\][/tex]
This is the solution to the equation [tex]\(2^{-x} = 10^{1+x}\)[/tex]. The value of [tex]\(x\)[/tex] that satisfies the equation is:
[tex]\[
\boxed{\frac{-\log(10)}{\log(20)}}
\][/tex]
