1. ## Umm...exponent/square root problem

Okay Im trying to find log 5^x =5*sq rt of 5. Thats log base 5 with exponent x equals 5 times the square root of five or, with a base of five, what exponent do I need to get 5 times the square root of 5? Now I know to solve for an exponent you take the natural logarithm of both sides and solve for x which is easy and I know that the square root is the same thing as a number raised to the 1/2 power but even though I know how to solve components of this problem I dont know how to solve the actual problem.

Okay let me try this: $\displaystyle \log_{5}x=5\sqrt5$
Yes thats it, and I need to solve for x.

Okay Im trying to find log 5^x =5*sq rt of 5. Thats log base 5 with exponent x equals 5 times the square root of five or, with a base of five, what exponent do I need to get 5 times the square root of 5? Now I know to solve for an exponent you take the natural logarithm of both sides and solve for x which is easy and I know that the square root is the same thing as a number raised to the 1/2 power but even though I know how to solve components of this problem I dont know how to solve the actual problem.

Okay let me try this: $\displaystyle \log_{5}x=5\sqrt5$
Yes thats it, and I need to solve for x.
the way you've written it ...

$\displaystyle \log_{5}x=5\sqrt{5}$

... translates to

$\displaystyle 5^{5\sqrt{5}} = x$

are you sure you don't mean

$\displaystyle \log_5(5\sqrt{5}) = x$

where ...

$\displaystyle 5^x = 5\sqrt{5}$ ???

3. No, I think I wrote it the right way and I think I figured it out:
$\displaystyle log_{5}x=5\sqrt{5}$
$\displaystyle \sqrt{5}$ changes to $\displaystyle 5^{1/2}$
Now I have $\displaystyle log_{5}x=5*5^{1/2}$
This is asking what exponent do I raise a base of 5 to in order to get $\displaystyle 5\sqrt{5}$ or $\displaystyle 5*5^{1/2}$
to solve this:
$\displaystyle 5^x= 5*5^{1/2}$
$\displaystyle ln5^x= ln(5*5^{1/2})$
$\displaystyle x(ln5)= ln(5*5^{1/2})$
$\displaystyle x= [ln(5*5^{1/2})]/ln5$
$\displaystyle x=ln11.18/ln5$
x=1.5

No, I think I wrote it the right way and I think I figured it out:
$\displaystyle log_{5}x=5\sqrt{5}$
$\displaystyle \sqrt{5}$ changes to $\displaystyle 5^{1/2}$
Now I have $\displaystyle log_{5}x=5*5^{1/2}$
This is asking what exponent do I raise a base of 5 to in order to get $\displaystyle 5\sqrt{5}$ or $\displaystyle 5*5^{1/2}$
to solve this:
$\displaystyle 5^x= 5*5^{1/2}$
$\displaystyle ln5^x= ln(5*5^{1/2})$
$\displaystyle x(ln5)= ln(5*5^{1/2})$
$\displaystyle x= [ln(5*5^{1/2})]/ln5$
$\displaystyle x=ln11.18/ln5$
x=1.5
you did not write the log equation correctly.

you worked it as ...

$\displaystyle \log_5(5\sqrt{5}) = x$

change to an exponential equation ...

$\displaystyle 5^x = 5\sqrt{5} = 5^1 \cdot 5^{\frac{1}{2}} = 5^{\frac{3}{2}}$

$\displaystyle x = \frac{3}{2}$

5. thanks a lot