# Logarithms

• October 8th 2012, 08:56 PM
Eraser147
Logarithms
Attachment 25119Click to enlarge.
• October 8th 2012, 09:10 PM
MarkFL
Re: Logarithms
What do you think is the best first step?
• October 8th 2012, 09:10 PM
chiro
Re: Logarithms
Hey Eraser147.

Try collecting the 10^(t) terms to one side and then use the relationship that log_10(10^t) = t where log_10(x) = log(x)/log(10) where log(x) is the natural logarithm.
• October 8th 2012, 09:46 PM
Eraser147
Re: Logarithms
I ended up with $10^t+1=.8*10^t+1.6$. I don't know where to go from there.
• October 8th 2012, 09:53 PM
chiro
Re: Logarithms
10^(t) - 0.8*10^(t) = 1.6 - 1 which means
0.2*10^(t) = 0.6 which means
10^(t) = 0.6/0.2 = 3.
• October 8th 2012, 10:12 PM
Eraser147
Re: Logarithms
Oh... THANKS A BUNCH. My error was not realizing I should subtract the 0.8*10^(t)
• October 9th 2012, 04:09 AM
Plato
Re: Logarithms
Quote:

Originally Posted by Eraser147

I find this whole thread off.

$10^t+7.9=1.1(10^t)+7.7$

$0.1(10^t)=0.2$

$10^t=2$

$t=\log_{10}(2)$