https://file.aiquickdraw.com/mathbot/file/1727398284086vsrymvso.jpg

The equation provided in the image is:

\[
D - D_0 = (D_1 - D_0) 10^{-kt}
\]

You are asked to solve for the variable \( t \). Let's solve it step by step.

### Step 1: Isolate the exponential term
First, divide both sides of the equation by \( (D_1 - D_0) \):

\[
\frac{D - D_0}{D_1 - D_0} = 10^{-kt}
\]

### Step 2: Take the logarithm of both sides
Since the right-hand side is an exponential, apply the logarithm (base 10) to both sides:

\[
\log_{10} \left( \frac{D - D_0}{D_1 - D_0} \right) = \log_{10} \left( 10^{-kt} \right)
\]

Using the logarithmic property \( \log_{10}(10^x) = x \), we simplify the right-hand side:

\[
\log_{10} \left( \frac{D - D_0}{D_1 - D_0} \right) = -kt
\]

### Step 3: Solve for \( t \)
Finally, solve for \( t \) by dividing both sides by \( -k \):

\[
t = -\frac{1}{k} \log_{10} \left( \frac{D - D_0}{D_1 - D_0} \right)
\]

Thus, the solution for \( t \) is:

\[
t = -\frac{1}{k} \log_{10} \left( \frac{D - D_0}{D_1 - D_0} \right)
\]

Would you like more details on any specific step, or have any further questions?

### 5 Relative Questions:
1. What is the significance of the constant \( k \) in this equation?
2. How would the solution change if the logarithm used was natural (base \( e \))?
3. Can this equation model any real-world phenomenon such as radioactive decay or population growth?
4. How does the ratio \( \frac{D - D_0}{D_1 - D_0} \) affect the time \( t \)?
5. How would you interpret negative values of \( t \) in this context?

### Tip:
Always check the units of the constants and variables in equations like this to ensure they are consistent before solving.

Solve for the indicated variable t in the equation: D - D_0 = (D_1 - D_0) 10^{-kt}

Learn how to solve for the variable t in the exponential decay equation D - D_0 = (D_1 - D_0) 10^{-kt} using logarithmic properties and algebraic manipulation. This problem is ideal for high school students in Grades 10-12 and covers key concepts in logarithms and exponential functions.

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