We know this equation a°=1.prove this?, Mathematics

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we know that log1 to any base =0 take antilog threfore a0 =1


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integration: if f(x)+f(x+1/2) =1 find limit 0 to 2, f(x)+f(x+1/2) =1 f(x...

f(x)+f(x+1/2) =1 f(x)=1-f(x+1/2) 0∫2f(x)dx=0∫21-f(x+1/2)dx 0∫2f(x)dx=2-0∫2f(x+1/2)dx take (x+1/2)=v dx=dv 0∫2f(v)dv=2-0∫2f(v)dv 2(0∫2f(v)dv)=2 0∫2f(v)dv=1 0∫2f(x)dx=1

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