marcelie
  • marcelie
Help please !!! Solve the following with a calculator. 2log (100^-3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
marcelie
  • marcelie
|dw:1444009782868:dw|
anonymous
  • anonymous
-12
marcelie
  • marcelie
how you get that

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anonymous
  • anonymous
Just put it into your calculator or google :)
marcelie
  • marcelie
lol z.. i wanna know know how solve it by hand.
Nnesha
  • Nnesha
apply the power rule
Nnesha
  • Nnesha
power rule \[\large\rm log_b x^y = y \log_b x\]
Nnesha
  • Nnesha
\[\huge\rm 2 \log (100^{-3})\] number at front of log becomes exponent of (100^{-3})\[\large\rm log (100^{-3})^2\] now multiply the exponents \[\rm (X^m)^n=x^{m \times n}\] and then negative exponent rule \[\large\rm x^{-m}=\frac{ 1 }{ x^m }\] then there are two ways to find final answer 1st) use change of base formula and then calculator or 2nd) set it equal to x to solve by hand
Nnesha
  • Nnesha
make sense ? take ur time to read that :=)
marcelie
  • marcelie
one second let me attach the picture
marcelie
  • marcelie
1 Attachment
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\)
Nnesha
  • Nnesha
gimme a sec i'll check it again
Nnesha
  • Nnesha
looks right
marcelie
  • marcelie
aiight got it :) do you know any website that has log formulas
Nnesha
  • Nnesha
quotient rule\[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\]
Nnesha
  • Nnesha
heheh change of base formula \[\huge\rm \log_b a = \frac{ \log a }{ \log b }\]
Nnesha
  • Nnesha
http://www.rapidtables.com/math/algebra/Logarithm.htm
Jhannybean
  • Jhannybean
\[\begin{align} 2\log(100^{-3}) &=2\log\left(\frac{1}{100^3}\right) \\& = 2[\log(1) -\log(100^3)] \\ &=2[0-\log(10^6)] \\&= 2[-6\log(10)] \\&= 2(-6(1))\\&= -12 \end{align} \]
Jhannybean
  • Jhannybean
That's another approach you could have used as well :P
marcelie
  • marcelie
ah i see it :) Thank you ladies :)
Jhannybean
  • Jhannybean
@Nnesha what do you think?
Nnesha
  • Nnesha
nice!
Nnesha
  • Nnesha
or after this if you don't want to use change of base formula set it equal to x \[\huge\rm log (100^{-6}) =x\] now change log to an exponential form log= log base 10 \[10^x =100^{-6}\] rewrite 100^{-16} in terms of base 10 \[10^x =10^{2 \times -6}\] bases are same you can cancel \[\cancel{10}^x =\cancel{10}^{2 \times -6}\] left with x= 2 times -6 = -12 :=)
Nnesha
  • Nnesha
|dw:1444012127376:dw| that's all you need to know for logs or u can visit that website :=)
Jhannybean
  • Jhannybean
\[\large \log_{10}(100^{-6}) = x\]\[10^{\large \log_{10}(100^{-6})} = 10^x\]\[\large 100^{-6} = 10^{x} \qquad \qquad \qquad 10^{-12} \equiv 100^{-6} \]\[\large 10^{-12} = 10^{x}\]\[\therefore -12 = x\]

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