# Thread: Solving simultaneously involving logs.

1. ## 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 ?

2. 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:

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

which yields $\displaystyle m \approx 0.09354...$

Re-substitute into [A] to get x.

3. 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.

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