#### victoria0212

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Here's it is: Write 323∙√3 as a exponent with the base 3. Use the exponent laws. No calculator.

- Thread starter victoria0212
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Here's it is: Write 323∙√3 as a exponent with the base 3. Use the exponent laws. No calculator.

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This is a very odd question. Is it copied correctly? Answer as written: \(323\cdot 3^{\tfrac{1}{2}}\)

Here's it is: Write 323∙√3 as a exponent with the base 3. Use the exponent laws. No calculator.

Maybe you mean

\(\displaystyle 323 \cdot \sqrt{3} = 3^{\left(\frac 1 2 + \log_3(323)\right)}\) ?

To see this

by definition of logs \(\displaystyle 3^{\log_3(323)} = 323\)

\(\displaystyle \sqrt{3} = 3^{\frac 1 2}\)

combining these we get the expression above.

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I have actually no idea. This is how it was written but I don't quite understand it.

Maybe you mean

\(\displaystyle 323 \cdot \sqrt{3} = 3^{\left(\frac 1 2 + \log_3(323)\right)}\) ?

To see this

by definition of logs \(\displaystyle 3^{\log_3(323)} = 323\)

\(\displaystyle \sqrt{3} = 3^{\frac 1 2}\)

combining these we get the expression above.

- Joined
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Now that I've been thinking about this for a while and done my research, I think I finally get it. Thanks!!I think it's just a multiplication problem that needs a product written as a power of 3.

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Could you share your answer so that a student in future can be helped?Now that I've been thinking about this for a while and done my research, I think I finally get it. Thanks!!