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Mathematics 43 Online
OpenStudy (anonymous):

Solve without calculator. 2^(5-x)= 6

OpenStudy (kc_kennylau):

ln both sides first

OpenStudy (kc_kennylau):

With potentially any base

OpenStudy (kc_kennylau):

\[\Large\begin{array}{rcl} 2^{5-x}&=&6\\ \ln2^{5-x}&=&\ln6\\ (5-x)\ln2&=&\ln6\\ 5\ln2-x\ln2&=&\ln6\\ x\ln2&=&5\ln2-\ln6\\ x&=&\frac{5\ln2-\ln6}{\ln2} \end{array}\]

OpenStudy (kc_kennylau):

\[\Large\begin{array}{rcl} 2^{5-x}&=&6\\ \log_22^{5-x}&=&\log_26\\ 5-x&=&\log_26\\ x&=&5-\log_26 \end{array}\]

OpenStudy (kc_kennylau):

Any step not understood?

OpenStudy (anonymous):

it says the answer is \[5-\frac{ \log6 }{\log2 }\]

OpenStudy (kc_kennylau):

That's just the same

OpenStudy (kc_kennylau):

I forgot to tell you one more formula: \[\Large\frac{\log_na}{\log_nb}=\log_ba\]

OpenStudy (anonymous):

ohh I'm really bad at these

OpenStudy (kc_kennylau):

With practice you'll excel in no time :)

OpenStudy (anonymous):

apparently i need more practice then i thought

OpenStudy (kc_kennylau):

Find log25/log5 without a calculator

OpenStudy (anonymous):

log5?

OpenStudy (kc_kennylau):

Nope

OpenStudy (kc_kennylau):

\[\frac{\log25}{\log5}=\log_525=2\]

OpenStudy (anonymous):

ugghh im never going to get it

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