Question 886288: log_2((x^2)+ 1)−log_4(x^2) = 1
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i get x = 1
start with log2(x^2 + 1) - log4(x^2) = 1
take log4(x^2) and set it equal to y
you get log4(x^2) = y if and only if 4^y = x^2
4^y is equivalent to 2^(2y) and you get:
2^(2y) = x^2
you get 2^(2y) = x^2 if and only if log2(x^2) = 2y
solve for y and you get:
y = log2^(x^2) / 2
since log4^(x^2) is also equal to y, you get:
log2^(x^2) / 2 = log4(x^2)
replace log4(x^2) with log2(x^2)/2 in your original equation and you get:
log2(x^2 + 1) - log2(x^2)/2 = 1
multiply both sides of this equation by 2 to get:
2*log2(x^2 + 1) - log2(x^2) = 2
since 2*log2(x^2 + 1) is equivalent to log2((x^2+1)^2), then your equation becomes log2((x^2+1)^2) - log2(x^2) = 2
since log(a) - log(b) = log(a/b), your equation becomes:
log2(((x^2+1)^2)/x^2) = 2
this is true if and only if 2^2 = ((x^2+1)^2/x^2)
simplify this and you will get:
4 = (x^4 + 2x^2 + 1) / x^2
multiply both sides of this equation by x^2 and you get:
4x^2 = x^4 + 2x^2 + 1
subtract 4x^2 from both sides of this equation and you get:
x^4 - 2x^2 + 1 = 0
let y = x^2 and the equation becomes:
y^2 - 2y + 1 = 0
factor this equation and you get:
(y-1)^2 = 0
solve for y and you get y = 1
since y = x^2, then you get x^2 = 1
solve for x and you get x = +/- sqrt(1) which becomes x = +/- 1
these should be your solutions.
plug them into your original equation to see if they're good.
your original equation is:
log2(x^2 + 1) - log4(x^2) = 1
log2(1^2 + 1) = log2(2) = 1
log2((-1)^2 + 1) = log2(2) = 1
log4(1^2) = log4(1) = 0
log4((-1)^2) = log4(1) = 0
your original equation of log2(x^2 + 1) - log4(x^2) = 1 becomes 1 - 0 = 1 which is true.
your solutions are x = 1 and x = -1
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