# Laws of Logarithms

log 1050 = log 73.3 + 0.75 log m

I tried this: log 1050 = log 73.3 m^0.75

1050/73.3=73.3(m^0.75)/73.3

m^0.75=14.3247

logm14.3247=0.75

How do you solve for m?

Using $m\log a=\log (a^m)$ and $\log b-\log c=\log\dfrac bc$ where all the logarithms remain defined unlike $2\log(-1)=\log(-1)^2=\log1=0$
We have $\log \left(m^{.75}\right)=\log\dfrac{1050}{73.3}=\log\dfrac{1050\cdot3}{220}$
$$\implies m^{.75}=\dfrac{315}{22}$$
$$\implies m=\left(\dfrac{315}{22}\right)^{\dfrac43}$$