Let’s start by analyzing the given equations for the mass of sodium remaining after [tex]\( y \)[/tex] years.
### Step 1: Given Equations
We are provided with two initial equations:
1. [tex]\( S = 800 \cdot (0.165)^y \)[/tex]
2. [tex]\( S = 800 \cdot (0.766)^{2.6/y} \)[/tex]
### Step 2: Equating the Expressions for [tex]\( S \)[/tex]
Since both equations represent the same physical quantity, we can set them equal to each other:
[tex]\[ 800 \cdot (0.165)^y = 800 \cdot (0.766)^{2.6/y} \][/tex]
### Step 3: Simplify the Equation
We can divide both sides of the equation by 800 to isolate the terms with the exponents:
[tex]\[ (0.165)^y = (0.766)^{2.6/y} \][/tex]
### Step 4: Apply Logarithms to Both Sides
To solve for [tex]\( y \)[/tex], apply logarithms to both sides. For simplicity, let's use the natural logarithm [tex]\( \ln \)[/tex]:
[tex]\[ \ln((0.165)^y) = \ln((0.766)^{2.6/y}) \][/tex]
### Step 5: Use Logarithmic Properties
Utilize the power rule of logarithms [tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]:
[tex]\[ y \cdot \ln(0.165) = \frac{2.6}{y} \cdot \ln(0.766) \][/tex]
### Step 6: Isolate the Variable [tex]\( y \)[/tex]
We can multiply both sides by [tex]\( y \)[/tex] to remove the fractional exponent:
[tex]\[ y^2 \cdot \ln(0.165) = 2.6 \cdot \ln(0.766) \][/tex]
### Step 7: Solve for [tex]\( y \)[/tex]
Rearrange the equation to isolate [tex]\( y^2 \)[/tex]:
[tex]\[ y^2 = \frac{2.6 \cdot \ln(0.766)}{\ln(0.165)} \][/tex]
Finally, solve for [tex]\( y \)[/tex] by taking the square root of both sides:
[tex]\[ y = \sqrt{\frac{2.6 \cdot \ln(0.766)}{\ln(0.165)}} \][/tex]
### Conclusion
The above steps lead us to an equation that relates [tex]\( y \)[/tex] with the logarithms of the given constants. To find the precise numerical value of [tex]\( y \)[/tex], you would evaluate the natural logarithms and complete the arithmetic.
However, from the equations and analysis provided, we now have the form of [tex]\( y \)[/tex]. Notice that our key point is recognizing how the equations match each other through exponent rules and logarithms, rather than calculating distinct numbers ourselves.
