4\left(2\right)^{x-3}+3=95  solve the exponential equation, express irrational solutions as decimals correct to the nearest thousandth.

To solve the exponential equation:

\[
4\left(2\right)^{x-3} + 3 = 95
\]

### Step 1: Isolate the exponential term
Subtract 3 from both sides:

\[
4 \cdot 2^{x-3} = 92
\]

### Step 2: Divide by 4
\[
2^{x-3} = 23
\]

### Step 3: Apply logarithms
To solve for \( x \), apply the logarithm to both sides. Using the natural logarithm (\( \ln \)), we get:

\[
\ln(2^{x-3}) = \ln(23)
\]

Using the power rule of logarithms (\( \ln(a^b) = b \ln(a) \)), this becomes:

\[
(x - 3) \ln(2) = \ln(23)
\]

### Step 4: Solve for \( x \)
Now, divide both sides by \( \ln(2) \):

\[
x - 3 = \frac{\ln(23)}{\ln(2)}
\]

Then, add 3 to both sides:

\[
x = 3 + \frac{\ln(23)}{\ln(2)}
\]

### Step 5: Calculate the result
Now, substitute the values for \( \ln(23) \) and \( \ln(2) \):

\[
x \approx 3 + \frac{3.1355}{0.6931}
\]

\[
x \approx 3 + 4.524
\]

\[
x \approx 7.524
\]

So, the solution to the equation, rounded to the nearest thousandth, is:

\[
x \approx 7.524
\]

---

Would you like a detailed explanation of logarithmic properties or any other part? 

Here are 5 additional questions you might find interesting:
1. How do you apply the change of base formula for logarithms?
2. What are the steps to solve exponential equations with different bases?
3. How do you use logarithmic functions to model real-world exponential growth?
4. What is the importance of isolating the exponential term in equations?
5. How do you solve equations involving exponential decay?

**Tip:** When solving exponential equations, isolating the exponential term first simplifies the use of logarithms.

4(2)^(x-3)+3=95 solve the exponential equation, express irrational solutions as decimals correct to the nearest thousandth.

Master solving exponential equations with this step-by-step solution for 4(2)^(x-3) + 3 = 95. Learn how to isolate the exponential term, apply logarithms, and approximate irrational solutions to the nearest thousandth. Suitable for high school students, Grades 10-12.

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