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OpenStudy (anonymous):
Solve without calculator.
2^(5-x)= 6
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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?
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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 :)
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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):
\[\frac{\log25}{\log5}=\log_525=2\]
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OpenStudy (anonymous):
ugghh im never going to get it
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