Mathematics
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OpenStudy (juan1857):
Solve lnx-ln6=0
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jhonyy9 (jhonyy9):
use
lna - lnb = ln(a/b)
OpenStudy (juan1857):
so ln(x/6)?
jhonyy9 (jhonyy9):
yes
OpenStudy (juan1857):
is there anymore
jhonyy9 (jhonyy9):
how you know the function logarithmic can being zero
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OpenStudy (juan1857):
i don't know
jhonyy9 (jhonyy9):
sorry wait please
jhonyy9 (jhonyy9):
lnx -ln6 =0
lnx =ln6
x=6
OpenStudy (zzr0ck3r):
correct
OpenStudy (juan1857):
oh so that's how you do it
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OpenStudy (zzr0ck3r):
well, you sort of skipped something
OpenStudy (zzr0ck3r):
\(\ln(6)-\ln(x)=0\implies \ln(6)=\ln(x) \implies e^{\ln(x)}=e^{\ln(x)}\implies x=6\)
OpenStudy (zzr0ck3r):
but we often skip that step, or argue that since \(\ln(x)\) is a 1-1 function, \(\ln(a)=\ln(b)\implies a=b\)
OpenStudy (zzr0ck3r):
but you could not do that with other functions like \(f(x) = x^2\) because it is true that \(f(2)=2^2=(-2)^2=f(-2)\) but \(2\ne -2\)
OpenStudy (zzr0ck3r):
Oh woops, I thought you asked this question @jhonyy9 . Do you understand @juan1857 ?
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OpenStudy (juan1857):
yeah sorta
jhonyy9 (jhonyy9):
ok. @zzr0ck3r ty
OpenStudy (juan1857):
so is it practically what @jhonyy9 only more complicated for me lol
jhonyy9 (jhonyy9):
yes you see it right juan
OpenStudy (zzr0ck3r):
as long as you get it. I just added one more step that often gets skipped and most teachers dont mind. I don't...
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OpenStudy (juan1857):
wait, on your example what happened to the lns,
OpenStudy (juan1857):
ok @zzr0ck3r
OpenStudy (juan1857):
thanks