1. ## solving exponent

$\displaystyle 3\sqrt5=5^{2x}$

I thought you take the sqrt and you get $\displaystyle 5^{1/3}=5^{2x}$

Then you get $\displaystyle 5^3=5^{2x}$ So now the bases are the same and you can now solve 3=2x. Divide 2 and you get $\displaystyle x=3/2$

The answer is not that. Where did I go wrong?

2. Originally Posted by Darkhrse99
$\displaystyle 3\sqrt5=5^{2x}$

I thought you take the sqrt and you get $\displaystyle 5^{1/3}=5^{2x}$

Then you get $\displaystyle 5^3=5^{2x}$ So now the bases are the same and you can now solve 3=2x. Divide 2 and you get $\displaystyle x=3/2$

The answer is not that. Where did I go wrong?
Hello,

Is it $\displaystyle \sqrt[3]{5}$ or $\displaystyle 3 \cdot \sqrt{5}$ ?

There is another problem : $\displaystyle 5^{1/3} \neq 5^3$

You have to work the same way, but with keeping 1/3.

1/3=2x

--> x=1/6

3. Originally Posted by Darkhrse99
$\displaystyle 3\sqrt5=5^{2x}$

I thought you take the sqrt and you get $\displaystyle 5^{1/3}=5^{2x}$

Then you get $\displaystyle 5^3=5^{2x}$ So now the bases are the same and you can now solve 3=2x. Divide 2 and you get $\displaystyle x=3/2$

The answer is not that. Where did I go wrong?
You're going to have to use logarithms to solve this one. But first note that $\displaystyle \sqrt{5} = 5^{0.5}$. So you can divide this out and then you get $\displaystyle 3 = 5^{(2x - 0.5)}$. Take logarithms of both sides:
$\displaystyle \ln 3 = \ln (5^{(2x - 0.5)})$

$\displaystyle \ln 3 = (2x - 0.5) \ln 5$

$\displaystyle \frac{\ln 3}{\ln 5} = 2x - 0.5$

$\displaystyle \frac{\ln 3}{\ln 5} + 0.5 = 2x$

$\displaystyle \frac{\ln 3}{2 \ln 5} + 0.25 = x$

4. Originally Posted by Moo
Hello,

Is it $\displaystyle \sqrt[3]{5}$ or $\displaystyle 3 \cdot \sqrt{5}$ ?

There is another problem : $\displaystyle 5^{1/3} \neq 5^3$

You have to work the same way, but with keeping 1/3.

1/3=2x

--> x=1/6
It's $\displaystyle \sqrt[3]{5}$.