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)

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- Sep 11th 2009, 07:23 AMChinnie15Logarithmic 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, 07:29 AMsongoku
Hi Chinnie15

Is this the question :

$\displaystyle \frac{2^2}{\log_5 x}=\frac{1}{16}$ ? - Sep 11th 2009, 07:35 AMChinnie15
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, 07:43 AMsongoku
$\displaystyle 2^{\frac{2}{\log_5 x}}=\frac{1}{16}$ ?

- Sep 11th 2009, 07:55 AMFinley
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:

Answer:

(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, 08:04 AMChinnie15
- Sep 11th 2009, 08:13 AMFinley
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)

$\displaystyle 2^{2/\log_{5} x}=1/2^4$

$\displaystyle 2^{2/\log_{5} x}=2^{-4}$

Bases are the same, so therefore the indices must be equal!

$\displaystyle 2/\log_{5} x=-4$

Solving for X:

$\displaystyle 2/-4=\log_{5} x$

$\displaystyle -1/2=\log_{5} x$

Rearranging the log into index form (Using log laws):

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

Simplifying:

$\displaystyle 1/\sqrt{5}=x$

Rationalizing the denominator

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

:) - Sep 11th 2009, 12:40 PMChinnie15
hehe, thank you so much! :) I can see it more clearly now. I wasn't exactly sure how 1/16 was becoming -4, but now I do. They shouldn't be allowed to give you problems this complex, lol.