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Mathematics 55 Online
OpenStudy (anonymous):

Is there a way to change the base of a logarithm without the change of base formula and without a calculator? I believe it has to do with base 10 representation?

OpenStudy (anonymous):

tell u the truth... i never use the cob formula...

OpenStudy (anonymous):

lol ok...it's crucial for me though when I'm not allowed to use a calculator...

OpenStudy (anonymous):

give an example of what u wanna change...

OpenStudy (anonymous):

Anything really...if there are any restrictions, then something within them?

OpenStudy (anonymous):

ok, how bout this... change \(\log_2 32=x \) to log base 10... first i change to exponential form: \(\large 2^x = 32 \) then take log base 10 of both sides... \(\large log2^x = log32\rightarrow xlog_2=log32 \) then isolate x.... is that what u mean?

OpenStudy (anonymous):

That doesn't look familiar...I saw someone do it before, but I don't recall...what I saw had powers of 10 in the denominators and things like that.

OpenStudy (anonymous):

oops that should be \(\large xlog2 \) not \(\large xlog_2 \)

OpenStudy (anonymous):

so you're actually looking for the value for logs without the aid of a calculator... is that what u mean?

OpenStudy (anonymous):

Um...I don't think so...like they literally changed the logarithm using something called base ten representation. It turned something like \(\log_3x \implies \log_5 something\)

OpenStudy (anonymous):

\(\large log_3 x = y\rightarrow 3^y=x \) take the log base 5 of both sides: \(\large log_5 (3^y)=log_5 x\rightarrow ylog_5 3=log_5 x\rightarrow y=\frac{log_5 x}{log_5 3} \) like that?

OpenStudy (anonymous):

I think that will work :) I'll ask my friend who did it when I get the chance. Thanks for the help though :D

OpenStudy (anonymous):

ok... i'm curious what u find out... pls post a "post" post of this when u get the answer u needed... yw...:)

OpenStudy (anonymous):

alright I will try to remember to do that :) (I don't see my friend until school stats again)

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