# Logarithm questions

• Mar 7th 2009, 10:13 AM
Ananya
Logarithm 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, 10:17 AM
Jhevon
Quote:

Originally Posted by Ananya
Hi ,

I am unable to solve the following questions of log:

1. If log[html]<sub>10</sub>[/html] 24 = x , log<

as you can see, the html code is not working for you. please format your questions better in the future.

If $\log_{10}24 = x$, then ...? post the exact instructions
• Mar 7th 2009, 10:23 AM
Ananya
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, 01:12 AM
Quote:

Originally Posted by Ananya
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

Hi Ananya(Hi)

$log_{10}48$

$
= log_{10}(24 \times 2)
$

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

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

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

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

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

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

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

• Mar 9th 2009, 02:42 AM
Ananya

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