Hi,

Can anyone help me to solve the following:

I have to express log[base 10]48 in terms of x, y and z . Given that

log[10]24= x , log[10]80=y and log[10]25 = z

Best Regards,

Ananya

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- Mar 7th 2009, 09:13 AMAnanyaLogarithm questions
Hi,

Can anyone help me to solve the following:

I have to express log[base 10]48 in terms of x, y and z . Given that

log[10]24= x , log[10]80=y and log[10]25 = z

Best Regards,

Ananya - Mar 7th 2009, 09:17 AMJhevon
- Mar 7th 2009, 09:23 AMAnanya
Hi,

Can anyone help me to solve the following:

I have to express log[base 10]48 in terms of x, y and z . Given that

log[10]24= x , log[10]80=y and log[10]25 = z

Best Regards,

Ananya - Mar 8th 2009, 12:12 AMADARSH
Hi Ananya(Hi)

$\displaystyle log_{10}48 $

$\displaystyle

= log_{10}(24 \times 2)

$

$\displaystyle =log_{10}(24)+log_{10}(2)$

$\displaystyle =log_{10}{(24)+log_{10}(\frac{2^4\times 5}{25^{1/2}})}^{1/4}$

$\displaystyle

=x+log_{10}{(\frac{2^4\times 5}{25^{1/2}})}^{1/4}$

$\displaystyle

=x+\frac{log_{10}(\frac{2^4\times 5}{25^{1/2}})}{4}$

$\displaystyle

=x+\frac{ log_{10} {(2^4\times 5)} - log_{10} 25^{1/2} }{4}$

$\displaystyle

=x+\frac{log_{10}{(80)}-z/2}{4}$

$\displaystyle

=x+\frac{y-z/2}{4}$

GO ahead - Mar 9th 2009, 01:42 AMAnanya
Hi Adarsh,

Thanks for the solution (Yes) .

I have found out another way (Cool) :

log[base 10] 48

=log[base 10] {(24*80*25)/1000}

= log[10] 24 + log[10] 80 + log[10] 25 - log[10] 1000

= x + y + z - 3

Pardon me for posting the same post twice (Blush) ...i was getting restless and i thought that thread with the same heading can have related posts.

Best Regards,

Ananya