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?
Thanks!
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?
Just because I love you
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
$\displaystyle x = \sqrt{5}/5$