Question

how can you solve this, without using a calculator

Log (0.5) = x

I know the answer is -0.3, but I would like to know how to do it without using a calculator ( since you can not convert 0.5 into a power of 10 )

Answer 1

log have base 10 \[\huge\rm log_{10} .5 =x\]
change this to exponential form

Answer 2

5 x 10^-1

Answer 3

yep and then?

Answer 4

10^-1 ?

Answer 5

lol what ?

Answer 6

0.5 is 5 x 10^1 is the power of 10 of 0.5 yeah, but since its 5 x 10^-1, and not 1.0 x 10^(- something), how can u do it by hand?

Answer 7

how did you get 10^-1
now i'm confused >.<
isn't that suppose to loolk like this

Answer 8

hmmm

Answer 9

oh, no i was just re writing the equation as log (5x10^-1) = x

Answer 10

@javawarrior what does "solve" mean in this case?

is not -3 btw

Answer 11

so what's the next step ?

Answer 12

or -0.3 either, since it gives a float

Answer 13

just solving for x

Answer 14

answer is -.301

Answer 15

yeah thats the answer, but I want to know how you could do it without using a calculator

Answer 16

solving for "x"? well, is sorta already done then

log(0.5) = x <----- is already solved, nothing to do further I'd say

Answer 17

\[\frac{ \log .5 }{ \log 10 }\]
that's who we get -.30
but she wants to know who t get answer without using calc ~.^

Answer 18

there's a exam question: working out the pH of a buffer, using the henderson-hassleback equation right, and the exam doesn't permit calculators.

Answer 19

so if they reckon you can do it without a calculator, was wondering how you could it lol

Answer 20

well..... the number I get is a float
so.... unless you'd end up with a fraction, you'd need a calculator or take a good while getting -0.30102999566398119521

Answer 21

the whole equation is

7.21 + log(0.5) = x

Answer 22

oh they said in the question that

log 2 = 0.30

Answer 23

maybe you're expected to leave it in log terms

Answer 24

heheh

Answer 25

hmmmm

Answer 26

is there anyway you can relate log 2 = 0.30 with log 0.5 = x ?

Answer 27

so... come on now...
as paul harvey would say "what's the rest of the story"?

Answer 28

http://www.mathpapa.com/algebra-calculator.html

Answer 29

hahaha sorry guys, just saw it

Answer 30

Here is an interesting site: http://www.science-site.net/logcalc.htm

Answer 31

this sounds like
what's the rabbit color?
ohh and the the rabbit is albino
ohh and the rabbit has a ferret parent
ohh and the rabbit is aquatic.. and so on

Answer 32

thanks jvatsim, going to start reading it now

Answer 33

so none of you guys have a clue ?

Answer 34

@javawarrior give us the full story, rather than the remnant

ad yes, I do see the relevance of log(2)

Answer 35

So if I am reading this right... the procedure is as follows:

log(0.5) = log(5 x 10^-1) = log5 + log(10^-1) = log 5 - 1.
Then log(5) = log(10/2) = log 10 - log 2 = 1 - 0.30 = 0.70.
So, log 5 - 1 = 0.70 - 1 = -0.30

Answer 36

solve for x

7.21 + log0.5 = x

where log2=0.30

Answer 37

or post a quick screenshot of the material, so we could see what you've got

Answer 38

k

Answer 39

I think jvatsim has got it, but is there a quicker way ? probably not right ?

Answer 40

thanks @jtvatsim

Answer 41

quicker way... memorize all tables of logarithms... :) but not practical

Answer 42

\(\bf 7.21 + log(0.5) = x\implies 7.21+log\left( \cfrac{1}{2}\right)=x
\\ \quad \\

{\color{brown}{ log(a)-log(b)\to log\left( \frac{a}{b} \right)
}}
\\ \quad \\
7.21+log\left( \cfrac{1}{2}\right)=x\implies 7.21+[log(1)-log(2)]=x
\\ \quad \\
7.21+[log_{10}(1)-log_{10}(2)]=x
\\ \quad \\
{\color{brown}{ log_{10}(1)\to \square \iff 10^\square =1\implies \square =0 }}
\\ \quad \\
7.21+[{\color{brown}{ 0}}-log_{10}(2)]=x\implies 7.21-log(2)=x\)

Answer 43

hint:
\[\log \left( {0.5} \right) = \log \left( {\frac{1}{2}} \right) = - \log 2\]