1. ## logarithm equation

can someone please write back to me as soon as possible with the answer and working to this question, its got me totaly baffled!!

2log21/x + 3log2x = 4

2. Hello,

Originally Posted by jammy14
can someone please write back to me as soon as possible with the answer and working to this question, its got me totaly baffled!!

2log21/x + 3log2x = 4
$\displaystyle \log \frac 1x=-\log x$

Therefore, the equation becomes :

$\displaystyle -2\log_2 x+3 \log_2 x=4$

$\displaystyle \log_2 x=4$

---> $\displaystyle \underbrace{2^{\log_2 x}}_{x}=2^4$

$\displaystyle \boxed{x=16}$

3. thankx but could you please explain how you did the last bit-

--->

im not so good at this!!

4. Originally Posted by jammy14
thankx but could you please explain how you did the last bit-

--->

im not so good at this!!
Ok

Basically, $\displaystyle \log_a x=b$ means that $\displaystyle x=a^b$

Or you can remember that $\displaystyle a^{\log_a x}=x$

These are basic rules

------------

For the latter part, $\displaystyle 2^4=16$

5. thankx