Solve [5^(x^2)/5^(2x)] = 125

[jonboy]

The problem is: \(\displaystyle \L \;\frac{5^{x^{2}}}{5^{2x}}\,=\,125\)

I though of using the rule: \(\displaystyle log_{b}\frac{m}{n}\,=\,log_b{m}\,-\,log_b{n}\)

So: \(\displaystyle \L \;log_{5}x^2\,-\,log_{5}2x\,=\,125\)

....Now bring the exponents to the front of the log:\(\displaystyle \L \;x^2(.7)\,-\,2x(.7)\,=\,125\)

\(\displaystyle \L \;\,.7x^2\,-\,1.4x\,=\,125\)

Am I doing this right so far?

 

[galactus]

Try it this way jonboy:

Use your log properties. You got a little discombobulated near your last step.

\(\displaystyle \L\\\frac{5^{x^{2}}}{5^{2x}}=125\)

\(\displaystyle \L\\log\left(\frac{5^{x^{2}}}{5^{2x}}\right)=log(125)\)

\(\displaystyle \L\\log(5^{x^{2}})-log(5^{2x})=log(5^{3})\)

\(\displaystyle \L\\x^{2}log(5)-2xlog(5)=3log(5)\)

\(\displaystyle \L\\log5(x^{2}-2x)=3log(5)\)

\(\displaystyle \L\\x^{2}-2x=3\)

 

[jonboy]

That's beautiful Galactus! Thanks I got the answer now.

 

[skeeter]

no logs necessary ...

\(\displaystyle \L \frac{5^{x^2}}{5^{2x}} = 125\)

\(\displaystyle \L 5^{x^2 - 2x} = 5^3\)

\(\displaystyle \L x^2 - 2x = 3\)

 

[galactus]

Of course, skeeter trumped me with his showin' off.

 

[jonboy]

Haha nice Skeeter.