Question 815145
Since {{{B=10log((I/I[0]))}}} then
{{{B[1]=10log((I[1]/I[0]))}}} and {{{B[2]=10log((I[2]/I[0]))}}}<br>
We want {{{B[1]}}} and {{{B[2]}}} in the same equation. So we will combine the two above by adding or subtracting them. I can see ahead that subtraction will make things a little easier so we will subtract the second equation from the first one:
{{{B[1]-B[2]=10log((I[1]/I[0]))-10log((I[2]/I[0]))}}}
Adding {{{B[2]}}} to each side:
{{{B[1]=B[2]+10log((I[1]/I[0]))-10log((I[2]/I[0]))}}}
As you can, the desired equation is starting to appear. We just need to manipulate the right side so that<ul><li>the I's are gone</li><li>The d's are there</li><li>And it looks like the desired equation. The I's and d's are related by the equation: {{{I = k/d^2}}}. So {{{I[1] = k/d[1]^2}}} and {{{I[2] = k/d[2]^2}}}. Substituting for {{{I[1]}}} and {{{I[2]}}} we get:
{{{B[1]=B[2]+10log(((k/d[1]^2)/I[0]))-10log(((k/d[2]^2)/I[0]))}}}
Most of the I's are gone and we have the d's. All that is left is to eliminate the {{{I[0]}}}'s and the k's and make the equation look like the one we want. We do want only one log so I combine the two we have next. Factoring out 10:
{{{B[1]=B[2]+10(log(((k/d[1]^2)/I[0]))-log(((k/d[2]^2)/I[0])))}}}
Now we can use the {{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}} property to combine the logs:
{{{B[1]=B[2]+10(log((((k/d[1]^2)/I[0])/((k/d[2]^2)/I[0]))))}}}
Rewriting the division of fractions as a multiplication by the reciprocal:
{{{B[1]=B[2]+10(log((((k/d[1]^2)/I[0])*(I[0]/(k/d[2]^2)))))}}}
...and the {{{I[0]}}}'s and the k's cancel out!
{{{B[1]=B[2]+10(log((((1/d[1]^2)/1)*(1/(1/d[2]^2)))))}}}
which simplifies to:
{{{B[1]=B[2]+10(log((d[2]^2/d[1]^2)))}}}
We're getting very close. Rewriting the fraction:
{{{B[1]=B[2]+10(log(((d[2]/d[1])^2)))}}}
Using the {{{log(a, (p^n)) = n*log(a, (p))}}} property:
{{{B[1]=B[2]+10(2log((d[2]/d[1])))}}}
Multiplying the 10 and 2:
{{{B[1]=B[2]+20log((d[2]/d[1]))}}}