Umm...exponent/square root problem

**mgrade2000** · February 21st 2010, 03:40 PM

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: $\log_{5}x=5\sqrt5$
Yes thats it, and I need to solve for x.

**skeeter** · February 21st 2010, 04:27 PM

Originally Posted by mgrade2000

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: $\log_{5}x=5\sqrt5$
Yes thats it, and I need to solve for x.

the way you've written it ...

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

... translates to

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

are you sure you don't mean

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

where ...

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

**mgrade2000** · February 21st 2010, 04:53 PM

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

**skeeter** · February 21st 2010, 05:10 PM

Originally Posted by mgrade2000

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

you did not write the log equation correctly.

you worked it as ...

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

change to an exponential equation ...

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

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

**mgrade2000** · February 23rd 2010, 06:03 AM

thanks a lot

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