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marcelie
 one year ago
Help please !!! Solve the following with a calculator.
2log (100^3)
marcelie
 one year ago
Help please !!! Solve the following with a calculator. 2log (100^3)

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marcelie
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444009782868:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just put it into your calculator or google :)

marcelie
 one year ago
Best ResponseYou've already chosen the best response.0lol z.. i wanna know know how solve it by hand.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3power rule \[\large\rm log_b x^y = y \log_b x\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3\[\huge\rm 2 \log (100^{3})\] 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
 one year ago
Best ResponseYou've already chosen the best response.3make sense ? take ur time to read that :=)

marcelie
 one year ago
Best ResponseYou've already chosen the best response.0one second let me attach the picture

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3\(\huge\color{green}{\checkmark}\)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3gimme a sec i'll check it again

marcelie
 one year ago
Best ResponseYou've already chosen the best response.0aiight got it :) do you know any website that has log formulas

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3quotient 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
 one year ago
Best ResponseYou've already chosen the best response.3heheh change of base formula \[\huge\rm \log_b a = \frac{ \log a }{ \log b }\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align} 2\log(100^{3}) &=2\log\left(\frac{1}{100^3}\right) \\& = 2[\log(1) \log(100^3)] \\ &=2[0\log(10^6)] \\&= 2[6\log(10)] \\&= 2(6(1))\\&= 12 \end{align} \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1That's another approach you could have used as well :P

marcelie
 one year ago
Best ResponseYou've already chosen the best response.0ah i see it :) Thank you ladies :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1@Nnesha what do you think?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.3or 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
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444012127376:dw that's all you need to know for logs or u can visit that website :=)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \log_{10}(100^{6}) = x\]\[10^{\large \log_{10}(100^{6})} = 10^x\]\[\large 100^{6} = 10^{x} \qquad \qquad \qquad 10^{12} \equiv 100^{6} \]\[\large 10^{12} = 10^{x}\]\[\therefore 12 = x\]
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