How do you find the domain of the function P(x) = 10^x^2 + log(1-2x)

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How do you find the domain of the function P(x) = 10^x^2 + log(1-2x)

Mathematics
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since there is a logarithm in it; you have to limit the domain of "x" to values that make (1-2x)>0. Logarithms never reach 0 nor do they go negative so those values are meaningless. 1-2x>0 1>0+2x 1>2x 1>x flip it the whole thing around to read it easier and we get: x<1
hit the wrong button lol x< (1/2)
so the domain is 1/2?

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the domain is {x|x<(1/2)} in other words. x can be any value that is less than (1/2) another notation would be (-inf,(1/2)) think about it this way. In order for (1-2x) to be 0 (bad number; BAD number) "x" would have to be (1/2). Anything bigger than this would make (-2)(+number) = a negative number greater than 1. For instance: 1 - 2(1) = -1. Cant have negatives in the log function!!! log(-1) is strictly forbidden and is punishable by.... by.... well, something horrible I am sure. So, anything equal to and over (1/2) has got to be removed from the domain. That leaves everything less than (1/2) to negative infinity that CAN be used.
Thanks! Do you do anything with 10 to the power of x^2?
10^(x^2) has no limitations. In other words, the input value of an exponential function can be any "real" number and it will output a valid answer. But, because 10^(x^2) and log(1-2x) are sharing the same values for "x". We have to let log(1-2x) control what we can use as inputs. A domain tells us what values we can use for inputs in a function. Domain = Inputs. Make sense?

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