Question

log2x−log4x=−2


how then? :/

Answer 1

\[\log _{2}x-\log _{4}x\]

Answer 2

=-2

Answer 3

\[\log _{2}(x \div 4) = \log _{2} (4x)\]

and hence i got x=16x lol

Answer 4

but that is wrong for sure x is not 0

Answer 5

the answer is 1 over 16 btw

Answer 6

i think you have to use change of base formula
\[\log_{b} x = \frac{\ln x}{\ln b} \]

Answer 7

Hmm i nvr did algorithm for a long time, not sure if im right.

But here it is,
Do a change of base.
\[\log_{2} x - \log_{4} x= -2\]
\[\frac{\log_{4}x}{\log_{4}2}-\log_{4}x = -2\]
\[\frac{\log_{4}x}{2} - \log_{4}x= -2\]
\[\log_{4} x- 2 \log_{4}x = -4\]
\[\log_{4} x = 4\]
\[4= x ^{4}\]
\[x=4^{\frac{1}{4}}\]
x= 1.414

Answer 8

well i didnt get the answer, hmm think i went wrong somewhr.

Answer 9

\[\log _{4}2 \] is not 2 lol i saw ur mistake

Answer 10

Oh my god. told you im rusty. i got the algorithm totally the wrong way. damn it

Answer 11

\[\rightarrow \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 4} = -2\]
ln(4) = ln(2^2) = 2*ln(2)
\[\rightarrow \frac{2\ln x - \ln x}{2 \ln 2} = -2\]
\[\ln x = -4*\ln 2\]
\[x = e^{-4 \ln 2} = 2^{-4} = \frac{1}{16}\]

Answer 12

that is complicating but i get it thanks :)

Answer 13

your welcome, tiaph was correct as well...you can change it to any base you like...i just like using natural logs :)