Logarithms: (log3^x)(logx^2x)(log2x^y)=logx^y squared

[cyberspace]

Ok, so I have this problem:

Solve for x and y:

(log3^x)(logx^2x)(log2x^y)=logx^y squared

I know by the product of three logs that log3^y = logx^y squared, but I don't know what steps to take after this.

Thanks for the help!

 

[tkhunny]

Re: Logarithms Help!

If only I could tell what you mean.

Is this: (log3^x)

This: \(\displaystyle \log_{3}(x)\)

Or this: \(\displaystyle \log(3^{x})\)

 

[o_O]

Re: Logarithms Help!

I assume this is your question:
\(\displaystyle log_{3}x \cdot log_{x}2x \cdot log_{2x}y = log_{x}y^{2}\)

And you gotten this as your last line:
\(\displaystyle log_{3}y = log_{x}y^{2}\)

This must mean you already know the property:
\(\displaystyle log_{a}b = \frac{log_{c}b}{log_{c}a}\)
where c can be any workable base.

So why don't we apply that to your last line?
\(\displaystyle \frac{logy}{log3} = \frac{logy^{2}}{logx}\)

\(\displaystyle \frac{logy}{log3} = 2\frac{logy}{logx}\) where \(\displaystyle logy^{2} = 2logy\)

Can you carry on from here? As you can see, the log y cancels out. What does that tell you about what y can be?