Calculating Logs: Solving Log_{10}(1.8)

[3trQN]

How do i calculate log_{10}(1.8) ?

I don't have a method for this kind of thing... I know that 1.8 = 10^{x}

Only thing i know to do is to use trial and error or possibly something akin to Hero's method for calculating roots.

Suggestions? My depth of logic is shallow indeed.

 

[mathman]

This doesn't answer your question, but why do you want to calculate it? There are tables to look up. Also any decent scientific hand calculator will have it as one of the functions.

There are series expressions for the natural log. To convert to the base 10, there is a constant multiplier - look it up.

 

[3trQN]

I would just like to know how :)

There are series expressions for the natural log. To convert to the base 10, there is a constant multiplier - look it up.

Of course, silly me...

 

[Stevedye56]

Plug it in the calculator, some serious sarcasm on my part because i know you want to know how to do it without one.

 

[murshid_islam]

expand the series of \ln\left(1+x\right) and then plug in 0.8 in place of x in that series. now you have the value of \ln\left(1.8). if you divide this by \ln\left(10), you will get the value of \log_{10} (1.8)

 

[3trQN]

Yup, i had a doh moment as soon as mathman mentioned "series expansion"...

Thx.

 

[mathman]

A somewhat more useful series for ln is ln((1+x)/(1-x)). You can get the ln of any positive number, since the series converges for 0<x<1.

 

[Gib Z]

mathman, sorri if I am mistaken as i usualli am, being 14 and ignorant, but wen u sed it converges for 0<x<1, did u mean for (1+x)/(1-x). if u didnt, then how dus that give u the ln and any positive number?

 

[mathman]

mathman, sorri if I am mistaken as i usualli am, being 14 and ignorant, but wen u sed it converges for 0<x<1, did u mean for (1+x)/(1-x). if u didnt, then how dus that give u the ln and any positive number?

My mistake: should be -1<x<1. For 0<x<1, you can get any u>1, while for -1<x<0, you can get 0<u<1.

The formula is easily solved for x to get x=(u-1)/(u+1), and x will be in the required range as long as u>0.