a) LogP = 2 + logx

b) LogM = 1.477 - x

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- Nov 21st 2008, 07:46 PMbickitrainerWrite the following equations without logarithms:
a) LogP = 2 + logx

b) LogM = 1.477 - x

Help! - Nov 21st 2008, 08:11 PMDavid24
- Nov 21st 2008, 09:57 PMearboth
I assume that the base of the logarithms is 10. If so:

$\displaystyle \log(P)=2+\log(x)~\implies~10^{\log(P)}=10^{2+\log (x)}~\implies~P=100\cdot x$

I assume that $\displaystyle 1.477 = \log(30)$ . If so:

$\displaystyle \log(M)=1.477-x~\implies~10^{\log(M)}=10^{1.477-x}~\implies~ M=\dfrac{10^{1.477}}{10^x}\implies~ M=\dfrac{30}{10^x}$