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
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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 : $\displaystyle -2\log_2 x+3 \log_2 x=4$ $\displaystyle \log_2 x=4$ ---> $\displaystyle \underbrace{2^{\log_2 x}}_{x}=2^4$ $\displaystyle \boxed{x=16}$
thankx but could you please explain how you did the last bit- ---> im not so good at this!!
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, $\displaystyle 2^4=16$
thankx
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