1. ## natural log help

I'm using this formula to analyze data, and don't quite understand this step... I just can't quite remember if ln works like this, but if somebody can just either confirm that this is correct, or if not, that would be great

the original equation is ln(B) = c +α ln(d)

I have c,
α, and d, and am trying to solve for B

so, can I solve with B = c + d^
α

thanks!

2. ## Re: natural log help

Originally Posted by forgetful
the original equation is ln(B) = c +α ln(d)
I have c,
α, and d, and am trying to solve for B
Well the answer is $B=d^a\cdot e^c.$

3. ## Re: natural log help

Thanks Plato, had a feeling there was something missing