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How calculate x from

2=-20 log10(1/1+x^2)

Guest Apr 5, 2017
 #1
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+1

Solve for x:
2 = -(20 log(x^2 + 1))/(log(10))

2 = -(20 log(x^2 + 1))/(log(10)) is equivalent to -(20 log(x^2 + 1))/(log(10)) = 2:
-(20 log(x^2 + 1))/(log(10)) = 2

Divide both sides by -20/(log(10)):
log(x^2 + 1) = -(log(10))/10

-(log(10))/10 = log(1/10^(1/10)):
log(x^2 + 1) = log(1/10^(1/10))

Cancel logarithms by taking exp of both sides:
x^2 + 1 = 1/10^(1/10)

Subtract 1 from both sides:
x^2 = 1/10^(1/10) - 1

Take the square root of both sides:
Answer: | x = sqrt(1/10^(1/10) - 1)        or           x = -sqrt(1/10^(1/10) - 1)

Guest Apr 5, 2017
 #2
avatar+87294 
+1

2 = -20 log10(1/1+x^2)    divide both sides by -20

 

-1/10   = log10(1/1+x^2)

 

This says that

 

10^(-1/10)  =  1 / [ 1 + x^2 ]       exponentiate both sides to -10

 

10  =   [ 1 / ( 1 + x^2)]^(-10)      and we can write

 

10  = [ (1 + x^2) ] ^(10)     take each side to the 1/10 power

 

10^(1/10)  = 1 + x^2       subtract 1 from both sides

 

10^(1/10) - 1   = x^2      take both roots

 

x =  ± sqrt [10^(1/10) - 1]  ≈  ± 0.50885

 

 

cool cool cool

CPhill  Apr 5, 2017

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