omgespit9q

2022-10-14

Laws of Logarithms

$$\mathrm{log}1050=\mathrm{log}73.3+0.75\mathrm{log}m$$

I tried this: $$\mathrm{log}1050=\mathrm{log}73.3{m}^{0.75}$$

$$1050/73.3=73.3({m}^{0.75})/73.3$$

$${m}^{0.75}=14.3247$$

logm14.3247=0.75

How do you solve for m?

$$\mathrm{log}1050=\mathrm{log}73.3+0.75\mathrm{log}m$$

I tried this: $$\mathrm{log}1050=\mathrm{log}73.3{m}^{0.75}$$

$$1050/73.3=73.3({m}^{0.75})/73.3$$

$${m}^{0.75}=14.3247$$

logm14.3247=0.75

How do you solve for m?

domwaheights0m

Beginner2022-10-15Added 11 answers

Using $m\mathrm{log}a=\mathrm{log}({a}^{m})$ and $\mathrm{log}b-\mathrm{log}c=\mathrm{log}{\displaystyle \frac{b}{c}}$ where all the logarithms remain defined unlike $2\mathrm{log}(-1)=\mathrm{log}(-1{)}^{2}=\mathrm{log}1=0$

We have $\mathrm{log}\left({m}^{.75}\right)=\mathrm{log}{\displaystyle \frac{1050}{73.3}}=\mathrm{log}{\displaystyle \frac{1050\cdot 3}{220}}$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{m}^{.75}={\displaystyle \frac{315}{22}}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}m={\left({\displaystyle \frac{315}{22}}\right)}^{{\displaystyle \frac{4}{3}}}$$

We have $\mathrm{log}\left({m}^{.75}\right)=\mathrm{log}{\displaystyle \frac{1050}{73.3}}=\mathrm{log}{\displaystyle \frac{1050\cdot 3}{220}}$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{m}^{.75}={\displaystyle \frac{315}{22}}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}m={\left({\displaystyle \frac{315}{22}}\right)}^{{\displaystyle \frac{4}{3}}}$$

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