solving exponent

**Darkhrse99** · May 6th 2008, 02:05 PM

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

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

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

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

**Moo** · May 6th 2008, 02:07 PM

Originally Posted by Darkhrse99

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

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

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

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

Hello,

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

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

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

1/3=2x

--> x=1/6

**icemanfan** · May 6th 2008, 02:11 PM

Originally Posted by Darkhrse99

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

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

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

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 $\sqrt{5} = 5^{0.5}$ . So you can divide this out and then you get $3 = 5^{(2x - 0.5)}$ . Take logarithms of both sides:
$\ln 3 = \ln (5^{(2x - 0.5)})$

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

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

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

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

**Darkhrse99** · May 6th 2008, 02:41 PM

Originally Posted by Moo

Hello,

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

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

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

1/3=2x

--> x=1/6

It's $\sqrt[3]{5}$ .

Thread: solving exponent

LinkBack

Thread Tools

Display

solving exponent

Similar Math Help Forum Discussions

Solving for X that is an exponent? Do have to use logs?

Solving for Variable in Exponent.

Solving for x....when x is both an exponent and denominator

solving for x (negative exponent)

solving for an exponent

Search Tags