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1/6^x+2=216 How do you solve for x?
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do I understrand the equation right? \[\frac {1}{6^x+2}=216 \]
log?
it's and expodentail x+2 instead of 6^x and then add 2 to all of it. And we haven't been taught the log method yet.
so this? \[\frac {1}{6^{x+2}}=216\]
\[\frac {1} {6^x} + 2 = 216\]
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I think it is: \[\Large \frac{ 1 }{ 6^{x+2} } = 216\]
\[\left( 1/6 \right)^{x+2}=216\]
216 = 6^3
3 different ways it can be read...and the 'question asker' isn't even here...
6^(-x-2) = 6^3 -x-2 = 3 x = -5
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