Q: How do youm solve logx equals 2?

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You can't. You need 2 equations to solve for 2 unknowns

You cannot solve it since it is not a true equation.

divide 20 by 10 which equals 2 the times 2 by 2 which equals 4

2y = 4 (divide both sides by 2 to solve) y = 2

x=2

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logx^2=2 2logx=2 logx=1 10^1=x x=10

logx^3logx^2log14 is 3logx2logxlog14 this equals 6 log14 (logx)^2 So for example, if y=6log14(logx)^2 the log x = square root of (y/6(log14))

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log(100x) can be written as log100 + logx. This =2+logx

-2 plus 5 equals +3

assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2

You can't. You need 2 equations to solve for 2 unknowns

You cannot solve it since it is not a true equation.

divide 20 by 10 which equals 2 the times 2 by 2 which equals 4

2y = 4 (divide both sides by 2 to solve) y = 2

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