$$ 10^{x=64} $$, then find the value of $$ 10^{\frac{x}{2}+1} $$

We are given:

\[
10^{x} = 64
\]

Our goal is to find the value of:

\[
10^{\frac{x}{2} + 1}
\]

### Step 1: Express $ x $ in terms of logarithms

To find $ x $, take the logarithm (base 10) of both sides of the equation:

\[
x = \log_{10}(64)
\]

Using the property $ \log_{10}(a^b) = b \log_{10}(a) $ and knowing that $ \log_{10}(10) = 1 $, we can evaluate this:

\[
x = \log_{10}(64) \approx 1.806
\]

### Step 2: Find $ \frac{x}{2} + 1 $

Now, calculate $ \frac{x}{2} + 1 $:

\[
\frac{x}{2} + 1 = \frac{1.806}{2} + 1 = 0.903 + 1 = 1.903
\]

### Step 3: Evaluate $ 10^{\frac{x}{2} + 1} $

Now, find the value of $ 10^{1.903} $:

\[
10^{1.903} \approx 80
\]

Thus, the value of $ 10^{\frac{x}{2} + 1} $ is approximately 80.

Would you like more details or further clarification?

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Here are five related questions to expand on this topic:
1. What is the value of $ 10^{x/3} $?
2. How do we apply logarithmic rules to solve exponential equations?
3. What is the general method for solving equations of the form $ a^x = b $?
4. Can you explain the relationship between exponents and logarithms?
5. How does the base of the logarithm affect the solution to such problems?

**Tip:** Understanding the properties of logarithms can make solving exponential equations much easier!

Solve the exponential equation 10^x = 64 and find the value of 10^{\frac{x}{2} + 1} using logarithms. The problem includes step-by-step calculations involving logarithmic rules and exponentiation. Suitable for students in Grades 9-11.

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