# Help with algebra

#### victoria0212

##### New member
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.

#### pka

##### Elite Member
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: $$323\cdot 3^{\tfrac{1}{2}}$$

#### firemath

##### Full Member
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 $$\displaystyle 3^{1/2}$$.

• victoria0212

#### firemath

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

#### Romsek

##### Full Member
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.

#### victoria0212

##### New member
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.

#### firemath

##### Full Member
I think it's just a multiplication problem that needs a product written as a power of 3.

• victoria0212

#### victoria0212

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

• firemath

#### Subhotosh Khan

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

• topsquark