Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (melissa_something):

Log Question:

OpenStudy (melissa_something):

\[\log _{10} 1/\sqrt{10}\]

OpenStudy (melissa_something):

Sqrt 10/ 10 is wrong :(

Nnesha (nnesha):

\[\log_{10} \frac{1}{\sqrt{10}}\] like this ?

OpenStudy (jhannybean):

Change of base: \(\log_a (x) =\dfrac{\log_b (x)}{\log_b (a)}\)

OpenStudy (melissa_something):

Oh yes! Lol

Nnesha (nnesha):

you can convert square root to an exponent if you want or use change of base formula

OpenStudy (jango_in_dtown):

log(a/b)=log a-log b so the given expression becomes log 1-log sqrt 10 =-log sqrt 10 =-log (10)^1/2=-1/2

OpenStudy (jhannybean):

Lets take \(x=\dfrac{1}{\sqrt{10}}\)

Nnesha (nnesha):

\[\sqrt{y} \] can be written as \[\rm y^\frac{ 1 }{ 2 }\] so \[\rm \log_{10} \frac{ 1 }{ (10)^\frac{ 1 }{ 2 } }\] move the (10)^/2 at the numerator

OpenStudy (melissa_something):

So it would be \[\log 1/\sqrt{10} /\log 10\] ??

OpenStudy (melissa_something):

Oh oh okay

OpenStudy (jhannybean):

\[\begin{align} \log_{10} \left(\frac{1}{\sqrt{10}}\right) &=\frac{\log\left(\frac{1}{\sqrt{10}}\right)}{\log(10)}\\& = \frac{\log(1)-\log(\sqrt{10})}{\log(10)} \\&=\frac{-\frac{1}{2}\log(10)}{\log(10)}\\&=-\frac{1}{2} \end{align}\]

OpenStudy (melissa_something):

Wow thanks for all the support :o

Nnesha (nnesha):

typo (10)^{1/2} remember the exponent rule when move base from the numerator to denominator sign of the exponent would change \[\huge\rm \frac{ 1 }{ x^{-m }}= x^m\]

OpenStudy (melissa_something):

Yes oh my gosh thank you @Nnesha

Nnesha (nnesha):

u already got the answer so i'm just gonna work itout \[\log_{10} 10^{-\frac{1}{2}}\] apply the power rule power rule \[\large\rm log_b x^y = y \log_b x\] \[-\frac{1}{2} \log_{10}10\] log_{10} 10 = 1 so left wth -1/2

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!