your familiarity with exponential notation is what is needed, since logs are exponents
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OpenStudy (anonymous):
thanks you both...
OpenStudy (anonymous):
if you recognize \(10^{-1}=0.1\) the you get the answer immediately. i
if you do not then it is going to be a problem
of course you could use a calculator
OpenStudy (anonymous):
oh okay...you see they gave me steps but they go all away around the world and its just confusing!!!
OpenStudy (anonymous):
here is the deal
if you want to solve
\[\log_b(x)=y\] for \(y\) then you write in equivalent exponential form as
\[b^y=x\] and then see if you know what \(y\) has to be
if so, you win
OpenStudy (anonymous):
for example if i want to solve
\[\log_2(32)=y\] i write
\[2^y=32\] and see if i know \(y\)
since \(2^5=32\) i know \(y=5\) i.e.
\[\log_2(32)=5\]
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