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How does e^(ln3)=3?
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\(a^{\log_a(x)}=x\)
omg yes! i remember this now! how would that apply to e^(2ln5)
By definition we have \(\large y=\log_a(x) \implies a^y=x\) so if we substitute in for y, we get \(\large x = a^y=a^{log_a(x)}\)
we also have this property \[\log(x^y)=y\log(x)\] so \[e^{2\ln(5) }=e^{\ln(2^5)}=2^5 = 32\]
wouldn't it be \[e ^{\ln(5^{2)}}\] ?
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yeah sorry
so 5^2 = 25
ahh ok i see it now :)
\(e^{2\ln(5) }=e^{\ln(5^2)}=5^2 = 25\)
long day....
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its fine! I understood :)
I appreciate your help!
np
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