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Mathematics 14 Online
OpenStudy (anonymous):

log (0,4^2)/20^4 =

OpenStudy (anonymous):

OpenStudy (anonymous):

\[\log(\frac{a}{b}) = \log(a) - \log(b)\]

OpenStudy (jhannybean):

Is that really a decimal point?

OpenStudy (anonymous):

yes it is..

OpenStudy (anonymous):

, and . on keyboard are close to each other, one can do mistake in that, simple to understand it..

OpenStudy (jhannybean):

Wth is \[\log\left(\frac{0.4^2}{20^4}\right)\]

OpenStudy (anonymous):

It is what you have written.. :P

OpenStudy (anonymous):

Not a function but a question to simplify it more..

OpenStudy (anonymous):

\[\log\left(\frac{0.4^2}{20^4}\right) = \log(0.4^2) - \log(20^4)\]

OpenStudy (anonymous):

And power rule for logs says that: \[\log(a^m) = m \cdot \log(a)\]

OpenStudy (jhannybean):

\[\log\left(\frac{a}{b}\right) = \log(a)-\log(b)\]\[\log\left(\frac{0.4^2}{20^4}\right) = 2\log(0.4) -4\log(20)\]

OpenStudy (anonymous):

@pain100 should we assume base as \(10\) or do you have something else in your pocket?

OpenStudy (anonymous):

First value will be negative...

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yes is reply to what?? first question or second one?

OpenStudy (jhannybean):

\[0.4= \frac{2}{5}\]\[\log(0.4^2) = 2\log\left(\frac{2}{5}\right)=2\cdot [\log(2)-\log(5)]\]

OpenStudy (anonymous):

@beans you are going now in depths.. :P

OpenStudy (jhannybean):

Lmao. Just playing around with the function.

OpenStudy (anonymous):

Who is this beans?? Oh my God,

OpenStudy (anonymous):

base 10

OpenStudy (jhannybean):

\[4\log(20)=4\log(5\cdot2^2) = 4\log(5)+2\log(2)\]

OpenStudy (anonymous):

Jhanny let us make it more simple..

OpenStudy (jhannybean):

I am!!

OpenStudy (jhannybean):

\[\log\left(\frac{0.4^2}{20^4}\right) = 2\log(0.4) -4\log(20)\]\[=2\log(2)-2\log(5)+4\log(5)+2\log(2)\]

OpenStudy (anonymous):

\(log_{10}(0.4) = -0.3979\) \(log_{10}(20) = 1.3010\) Like this.. :P

OpenStudy (jhannybean):

Noooo

OpenStudy (jhannybean):

I think mine is better :-)

OpenStudy (anonymous):

But finally you have to evaluate it..

OpenStudy (jhannybean):

Ik. But still!

OpenStudy (perl):

in european countries 0,4 = 0.4

OpenStudy (jhannybean):

\[\implies 2\log(2)-2\log(5)+4\log(5)+2\log(2)\]\[=4\log(2)-2\log(5) = \log(2^4)-\log(5^2)=\frac{\log(2^4)}{\log(5^2)}\]

OpenStudy (anonymous):

Suppose you have options like : a) -6 b) -12 c) None.. I got \(-5.9999\) which is close to \(-6\).. :P

OpenStudy (jhannybean):

Hah!

OpenStudy (anonymous):

Here is.. :P

OpenStudy (anonymous):

umm my friend says the answer is -6

OpenStudy (anonymous):

Who is your that friend?

OpenStudy (anonymous):

OpenStudy (anonymous):

is this true?

OpenStudy (jhannybean):

Oh I see where I messed up, lol.

OpenStudy (anonymous):

@Jhannybean again look at \(4 \log(20)\) evaluation you have done above..

OpenStudy (jhannybean):

Yeah ik ik,.. I see it,lol.

OpenStudy (anonymous):

\[4 \log(20) = 4 \log(5) + 4 \log(4)\]

OpenStudy (jhannybean):

Or... \[4\log(5)+8\log(2)\]

OpenStudy (jhannybean):

\[\log_{10}\left(\frac{0.4^2}{20^4}\right) = 2\log(0.4) -4\log(20)\]\[=2\log(2)-2\log(5)-4\log(5)-8\log(2)\]\[=-6\log(2)-6\log(5)\]\[=-6\]

OpenStudy (dan815):

umm

OpenStudy (anonymous):

Where @Jhannybean ?? No you are good everywhere. :P

OpenStudy (anonymous):

\[-6(\log(2) + \log(5))\]

OpenStudy (anonymous):

= -6 ??

OpenStudy (jhannybean):

Yes, plug it into your calculator lol

OpenStudy (anonymous):

I got -5. something.

OpenStudy (dan815):

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