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OpenStudy (anonymous):
Solve for t.
2ln|t + 4| = 1
Please & thank you :)
Absolute value t + 4 btw if you couldn't tell!
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OpenStudy (anonymous):
first use this property
\[aln(b)=ln(b^a)\]
OpenStudy (anonymous):
so just ln|t+4|^2.
OpenStudy (anonymous):
alright now use
\[e^{ln(b^a)}=b^a\]
OpenStudy (anonymous):
but remember if you make e the base on the left side you must do it on the right side
OpenStudy (anonymous):
What would the answer be...this is one of the problems on my AP calc summer assignment & I don't remember this from precalc
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OpenStudy (anonymous):
well you'd get |t+4|^2=e^1
OpenStudy (anonymous):
-2.35?
OpenStudy (anonymous):
jk....
OpenStudy (anonymous):
not quite take the square rt of both
OpenStudy (anonymous):
I did, so it would be |t+4| = √e
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OpenStudy (anonymous):
\[|t+4|=\sqrt{e}\]
OpenStudy (anonymous):
Then subtract 4
OpenStudy (anonymous):
yes however you have the variable t so you should be + -
OpenStudy (anonymous):
Okay, that's because of the absolute value right?
OpenStudy (anonymous):
wait it will always be positive neverm
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OpenStudy (anonymous):
Okay lol thank you
OpenStudy (anonymous):
so basically
t+4=-sqrt{e}
and
t+4=sqrt{e}
OpenStudy (anonymous):
Thanks again!
OpenStudy (anonymous):
\[\ln|t+4|=\frac{1}{2}\]
\[|t+4|=e^{\frac{1}{2}}\]
\[t+4=e^{\frac{1}{2}}\] or
\[t+4=-e^{\frac{1}{2}}\]
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