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