Why Do Logarithmic Graphs Vary Despite Similar Equations?

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The graphs of y=logx+log2x and y=log2x^2 appear different due to their domains. The first equation is undefined for x<0 because logarithms only output real numbers for positive inputs, while the second equation remains defined for all x due to the squaring of x. Both graphs are identical for x>0, but the first graph becomes undefined for negative values, whereas the second graph extends to negative x as an even function. Simplifying y=logx+log2x to y=log2x^2 does not change the fact that both functions require x>0 for real results. Thus, the key difference lies in the domain restrictions of the logarithmic functions involved.
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I graphed y=logx+log2x and y=log2x^2 and the graphs came out a lot different. However when you simplify y=logx+log2x you actually get y=log2x^2! Any ideas about why the graphs are so different yet the equations are pretty much the same? thanx
 
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Well notice that log(a) outputs real numbers only if a>0.

With that in mind, when x<0, in your first equation the inputs to the logs are negative so that function is not defined if you are graphing real numbers. The second equation is defined for x<0 because x2 yields a positive value.

If you look at the two graphs they are identical when x>0. When x<0, the first graph is undefined while the second graph is just an even extension of when x>0.Just as a note, when you are dealing with the function y=log(x)+log(2x), the domain that yields a real result is x>0. If you simplify the expression to y=log(2x2) then you have to remember that your domain is still x>0. In that sense the two functions are identical.
 
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In other words, log a+ log b= log ab if and only if log a and log b both exist- that is, as long as a and b are both positive. If a and b are both negative, log ab exists even though log a and log b do not.
 
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