Logarithm questions

**Ananya** · March 7th 2009, 09:13 AM

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

**Jhevon** · March 7th 2009, 09:17 AM

Originally Posted by Ananya

Hi ,

I am unable to solve the following questions of log:

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

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

what is your question?

If $\log_{10}24 = x$ , then ...? post the exact instructions

**Ananya** · March 7th 2009, 09:23 AM

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

**ADARSH** · March 8th 2009, 12:12 AM

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

$log_{10}48$

$ = 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}$

GO ahead

**Ananya** · March 9th 2009, 01:42 AM

Hi Adarsh,

Thanks for the solution

.

I have found out another way

:

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

...i was getting restless and i thought that thread with the same heading can have related posts.

Best Regards,
Ananya

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