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Mathematics 13 Online
OpenStudy (anonymous):

Evaluate. log10 0.1 =

OpenStudy (shubhamsrg):

what is 1/0.1?

OpenStudy (anonymous):

\[\log _{10}0.1\]

OpenStudy (shubhamsrg):

tell me this : value of 1/0.1 ??

OpenStudy (anonymous):

10

OpenStudy (shubhamsrg):

so we know 1/0.1 = 10 can we say that 1/10 = 0.1 ?

OpenStudy (shubhamsrg):

conversely*

OpenStudy (shubhamsrg):

hmmm..

OpenStudy (anonymous):

\[\log_b(x)=y\iff b^y=x\] so your job is to solve \[10^{y}=0.1\] for \(y\)

OpenStudy (anonymous):

\[\log_{10}0.1 = \log_{10}10^{-1} = -1 \log_{10}10 = -1\]

OpenStudy (anonymous):

your familiarity with exponential notation is what is needed, since logs are exponents

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. \[\log_2(32)=5\]

OpenStudy (anonymous):

okay I see it

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