Help with algebra

victoria0212

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Feb 13, 2020
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Can someone please show and explain how to solve this? I'm not sure if I'm doing it right.
Here's it is: Write 323∙√3 as a exponent with the base 3. Use the exponent laws. No calculator.
 
Can someone please show and explain how to solve this? I'm not sure if I'm doing it right.
Here's it is: Write 323∙√3 as a exponent with the base 3. Use the exponent laws. No calculator.
This is a very odd question. Is it copied correctly? Answer as written: \(323\cdot 3^{\tfrac{1}{2}}\)
 
Hello, and welcome to Free Math Help!

Do you know your exponent laws?

Here is one hint: Another way to write the square root of 3 is [MATH]3^{1/2}[/MATH].
 
Your wording is a bit confusing.

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.
 
Your wording is a bit confusing.

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