# Solving simultaneously involving logs.

• Jul 18th 2009, 11:23 PM
smmmc
Solving simultaneously involving logs.
1) 52=x*10^m
2) 80=x*10^3m

my attempt

from 1) rearrange to get x=52/10^m

then sub into 2)
so
80=52/10^m * 10^3m

so
log 10(80) = log10(52) - log10(10^m) + log10(10^3m)
thenn i get
80=52 - m +3m

Help, Thank You. im unsure abou the log 10(80) part i think it should be 10^80 ?
• Jul 19th 2009, 12:01 AM
earboth
Quote:

Originally Posted by smmmc
1) 52=x*10^m
2) 80=x*10^3m

my attempt

from 1) rearrange to get x=52/10^m [A]

then sub into 2)
so
80=52/10^m * 10^3m

so
log 10(80) = log10(52) - log10(10^m) + log10(10^3m)
...

You get:

$\log(80)-\log(52)=2m~\implies~m=\dfrac12(\log(80)-\log(52))$

which yields $m \approx 0.09354...$

Re-substitute into [A] to get x.
• Jul 19th 2009, 04:51 AM
HallsofIvy
Substitution always works but, with a problem like this, I think the simplest way to eliminate one unknown variable is to divide one equation by the other.

$\frac{80}{52}= \frac{x10^{3m}}{x10^m}= 10^{2m}$
and we have eliminated x. Now $2m= log\left(\frac{80}{52}\right)= log(80)- log(52)$