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 Post subject: Teslim.
PostPosted: Fri, 13 Jan 2012 20:04:23 UTC 
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Hello.

$\text{If:}\;\;\;\;2^a=81\;\;\;\text{and}\;\;\;\;2^b=3\;\;\;\;\;\;\text{Find:}\;\frac{a}{b}

$A)\;9

$B)\;1

$C)\;2

$D)\;4

I have tried nothing, because I don't know what to try!

$\text{I used computer but like always:}

$2^a=81

$\text{log}\;(2^a)=\text{log}\;(81)

$a\;\text{log}\;(2)=\text{log}\;(81)

$a=\frac{\text{log}\;(81)}{\text{log}\;(2)}

$a=\text{log}_2(81)

$a=4\;\text{log}_2(3)

a\;\approx\;6.3398500028846

And pretty much the same with the second one too:

$b\;\approx\;1.5849625007212

$\text{therefore:}\;\frac{a}{b}\;\approx\;3.9999999999999

But I would really like to know how to solve it without calculator!


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 Post subject: Re: Teslim.
PostPosted: Fri, 13 Jan 2012 20:48:44 UTC 
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kreshnik wrote:
Hello.

$\text{If:}\;\;\;\;2^a=81\;\;\;\text{and}\;\;\;\;2^b=3\;\;\;\;\;\;\text{Find:}\;\frac{a}{b}

$A)\;9

$B)\;1

$C)\;2

$D)\;4

I have tried nothing, because I don't know what to try!

$\text{I used computer but like always:}

$2^a=81

$\text{log}\;(2^a)=\text{log}\;(81)

$a\;\text{log}\;(2)=\text{log}\;(81)

$a=\frac{\text{log}\;(81)}{\text{log}\;(2)}

$a=\text{log}_2(81)

$a=4\;\text{log}_2(3)

a\;\approx\;6.3398500028846

And pretty much the same with the second one too:

$b\;\approx\;1.5849625007212

$\text{therefore:}\;\frac{a}{b}\;\approx\;3.9999999999999

But I would really like to know how to solve it without calculator!


Taking logs of both sides in base 3, you get:

a\log_3(2)=4
b\log_3(2)=1

Now do division:

{a\log_3(2)\over b\log_3(2)}={4\over 1}. Now cancel like terms.

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 Post subject: Re: Teslim.
PostPosted: Fri, 13 Jan 2012 21:09:06 UTC 
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Shadow wrote:
Taking logs of both sides in base 3, you get:

a\log_3(2)=4
b\log_3(2)=1

Now do division:

{a\log_3(2)\over b\log_3(2)}={4\over 1}. Now cancel like terms.


Thank you very much. :D


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 Post subject: Re: Teslim.
PostPosted: Fri, 13 Jan 2012 22:44:06 UTC 
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kreshnik wrote:
Shadow wrote:
Taking logs of both sides in base 3, you get:

a\log_3(2)=4
b\log_3(2)=1

Now do division:

{a\log_3(2)\over b\log_3(2)}={4\over 1}. Now cancel like terms.


Thank you very much. :D


Ooh! Alternatively, you can go from what you did and use the change of base formula:

\log_c(d)={\log d\over\log c}

Your work shows a={\log 81\over\log 2} and b={\log 3\over\log 2}, then you get {a\over b}={\log 81\over\log 3}=\log_3(81)

and one last way you could do it from the way you started is to use the log property \log(a^x)=x\log a so that you can avoid the change of base formula and just get ${a\over b}={\log(3^4)\over\log 3}={4\log 3\over\log 3}=4

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