Very silly question on whether the domain of ##log_{10}(x²)## = ##2log_{10}(x)##

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The discussion centers on finding the x-intercepts of the function log_{10}(x²) and its simplified form 2log_{10}(x). The original function yields x-intercepts at x = -1 and x = 1, while the simplified version only gives x = 1. It is clarified that the correct simplification should be 2log_{10}(|x|), which retains the intercepts of the original function. The conclusion emphasizes that both forms ultimately have the same x-intercepts when considering the absolute value.
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So say I have to find the x intercept of this function $$log_{10}(x²)$$ I get x={-1,1}.
But if I try to find the x intercept of this same function after simplifying I get $$2log_{10} (x)$$ I get x={1}
 
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tellmesomething said:
Homework Statement: Title
Relevant Equations: Title

So say I have to find the x intercept of this function $$log_{10}(x²)$$ I get x={-1,1}.
But if I try to find the x intercept of this ## same function## after simplifying I get $$2log_{10} (x)$$ I get x={1}
After simplifying, you get ##2log_{10} (|x|)## rather than ##2log_{10} (x)##, which has the same intercept as the original function.
 
Hill said:
After simplifying, you get ##2log_{10} (|x|)## rather than ##2log_{10} (x)##, which has the same intercept as the original function.
Okay then :-)
 
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