# Logarithmic equation

• Sep 11th 2009, 08:23 AM
Chinnie15
Logarithmic equation
Solve for x: 2^2/log_5 X=1/16

I know how to do the more basic ones, but this one is towards the end of the chapter and is very confusing. How would I go about doing this? (Worried)

Thanks! (Wink)
• Sep 11th 2009, 08:29 AM
songoku
Hi Chinnie15

Is this the question :

$\frac{2^2}{\log_5 x}=\frac{1}{16}$ ?
• Sep 11th 2009, 08:35 AM
Chinnie15
No, the 2/log_5 X is all in the superscript. The two stands alone with all that in it's exponent. Sorry if it was a little confusing. :(
• Sep 11th 2009, 08:43 AM
songoku
$2^{\frac{2}{\log_5 x}}=\frac{1}{16}$ ?
• Sep 11th 2009, 08:55 AM
Finley
The quick answer (without mathematical syntax):

2^(2/log_5 X)=1/16

2^(2/log_5 X) = 1/2^4

2^(2/log_5 X) = 2^-4

2/log_5 X = -4

2/-4 = log_5 X

X = 5^(-1/2)

= 1/(5^1/2)

Rationalized, this becomes:
(5^1/2)/5

Essentially, what you need to do is have the same bases across the board! The first thing I considered was the 16, which could be simplified to base 2 (a base which is common on the left hand side). After the bases are the same on both sides of the equation we can simply acknowledge that both indices must be equal and solve from there!

Make sense?
• Sep 11th 2009, 09:04 AM
Chinnie15
Thanks for your help! I do find it a little confusing without the syntax, I'm still not sure where you are getting some of these numbers from?

Quote:

Originally Posted by songoku
$2^{\frac{2}{\log_5 x}}=\frac{1}{16}$ ?

Yes, that's it :)
• Sep 11th 2009, 09:13 AM
Finley
Just because I love you :)

http://www.mathhelpforum.com/math-he...c708bfc5-1.gif
Make the bases the same!! (2 is the lowest base evident, so we'll compound 16 to match)
$2^{2/\log_{5} x}=1/2^4$
$2^{2/\log_{5} x}=2^{-4}$

Bases are the same, so therefore the indices must be equal!
$2/\log_{5} x=-4$

Solving for X:
$2/-4=\log_{5} x$
$-1/2=\log_{5} x$

$5^{-1/2}=x$
$1/\sqrt{5}=x$
$x = \sqrt{5}/5$