Solve the logarithmic equation below.
\[2 \log _{4}(5-x) - \log _{4}2=\log _{4}50\] That is the problem.
@satellite73 @jdoe0001 @cwrw238
I know \[\log_{4}2=.4\]
actually .5 but no matter
And \[\log_{4}50=\frac{ \log 50 }{ \log 4 }\]
Oops that was a typo..
you can write \[\log_4((5-x)^2) -.5=\log_4(50)\] if you like
Ok.
but really lets just leave the \[\log_4(2)\] alone and add it to the right \[\log_4((5-x)^2)=\log_4(50)+\log_4(2)\]
or \(\bf {\color{brown}{ 2}} \log _{4}(5-x) - \log _{4}2 \\ \quad \\ \log _{4}(5-x)^{\color{brown}{ 2}} - \log _{4}2=\log _{4}50\implies log_4\left(\cfrac{(5-x)^2}{2}\right)=log _{4}50\) and you could pretty much see the equation there
that tells you \[\log_4((5-x)^2)=\log_4(100)\] making \[(x-5)^2=100\] and that is real easy to solve
It was that easy.. I feel so dumb lol
you can just about solve that one in your head either \[5-x=10\\ x=-5\] or \[5-x=-10\\ x=15\]
check that the negative solution probably doesn't work in the original equation so toss that one out
Thank you so much @satellite73 @jdoe0001 :)
yw
Actually -5 was the answer not 15 but thanks.
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