https://file.aiquickdraw.com/mathbot/file/1727089160298rv15fjbi.jpg

The equation shown in the image is:

\[
4^{x-3} = \sqrt{8^{x-5}}
\]

Let's solve this step by step.

### Step 1: Simplify the right-hand side
The square root can be rewritten as a fractional exponent:
\[
\sqrt{8^{x-5}} = (8^{x-5})^{1/2} = 8^{\frac{x-5}{2}}
\]

So the equation becomes:
\[
4^{x-3} = 8^{\frac{x-5}{2}}
\]

### Step 2: Rewrite both sides with the same base
We can express both 4 and 8 as powers of 2:
\[
4 = 2^2 \quad \text{and} \quad 8 = 2^3
\]

Thus:
\[
4^{x-3} = (2^2)^{x-3} = 2^{2(x-3)} = 2^{2x - 6}
\]
\[
8^{\frac{x-5}{2}} = (2^3)^{\frac{x-5}{2}} = 2^{\frac{3(x-5)}{2}} = 2^{\frac{3x - 15}{2}}
\]

Now the equation is:
\[
2^{2x - 6} = 2^{\frac{3x - 15}{2}}
\]

### Step 3: Set the exponents equal
Since the bases are the same, we can equate the exponents:
\[
2x - 6 = \frac{3x - 15}{2}
\]

### Step 4: Solve for \(x\)
Multiply both sides of the equation by 2 to eliminate the fraction:
\[
2(2x - 6) = 3x - 15
\]
\[
4x - 12 = 3x - 15
\]

Subtract \(3x\) from both sides:
\[
x - 12 = -15
\]

Add 12 to both sides:
\[
x = -3
\]

### Conclusion:
The solution to the equation is \(x = -3\), which matches the first option in the list.

Would you like further details or have any questions?

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Here are 5 related questions for further exploration:
1. How do you convert a square root into a fractional exponent?
2. What is the general method for solving exponential equations?
3. How do you solve equations with different powers of the same base?
4. What happens if the equation had a different base, such as base 5 or base 10?
5. Can you solve an exponential equation graphically?

**Tip:** Always try to express terms with the same base when solving exponential equations, as it simplifies the comparison between exponents.

penyelesaian dari 4^(x-3) = sqrt(8^(x-5)) adalah ...

This page provides a detailed step-by-step solution to the exponential equation 4^(x-3) = sqrt(8^(x-5)). Learn how to simplify roots, convert bases to powers of 2, and equate exponents. Ideal for students in Grades 10-12.

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