Another Question of solving in Logarithm problem

G

Guest

Guest
Hi,

Suppose I have this problem and I have to solve for "r", and I use a calculator:

log 0.1 = 10 log (1 - r)

How do I solve for "r"?

If I do this:

log 0.1 / 10 = log (1 - r)

I can't put it in the calculator of course...

One of my friends said that I can do this:

log 0.1 / 10 = log (1 - r)

and then

log 0.1 / log 10 = (1 - r)

and finally

(log 0.1 / log 10) - 1 = r

I'm not even sure if you can do that.

I'm stuck.

Thanks.
 
By a TI-89 and you can solve for r easily.

But to do it the 'old-fashioned' way:

\(\displaystyle \L\\log(\frac{1}{10})=10log(1-r)\)

\(\displaystyle \L\\\frac{-1}{10}=log(1-r)\)

\(\displaystyle \L\\10^{\frac{-1}{10}}=1-r\)

\(\displaystyle \L\\r=1-10^{\frac{-1}{10}}\approx{0.205671765276}\)
 
galactus said:
By a TI-89 and you can solve for r easily.

But to do it the 'old-fashioned' way:

\(\displaystyle \L\\log(\frac{1}{10})=10log(1-r)\)

\(\displaystyle \L\\\frac{-1}{10}=log(1-r)\)

\(\displaystyle \L\\10^{\frac{-1}{10}}=1-r\)

\(\displaystyle \L\\r=1-10^{\frac{-1}{10}}\approx{0.205671765276}\)
Click to expand...

Wait why and how did you un-logged the fraction and made it negative? Where did the 10 on the right side go?

BTW we only use Ti-84's

Thanks!
 
log(1/10) = -log(10)

If log(a) = 10*B, then (1/10)log(a) = B

(1/10)log(a) = log(a^(1/10))
 
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