SOLUTION: Plz help solve for x if 2^(x+1) - 6^x = 2^(x-1), 3^x + 3^-x = 4 & dis 1 plz 2logbase3 (x+2)=2+logbase3 x?
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Question 751134: Plz help solve for x if 2^(x+1) - 6^x = 2^(x-1), 3^x + 3^-x = 4 & dis 1 plz 2logbase3 (x+2)=2+logbase3 x?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Take the log of both sides, any base...
\ =\ \ln\left(\frac{3}{2}\right))
\ =\ \ln\left(\frac{3}{2}\right))
}{\ln\left(3\right)})
Which is the exact answer. Use your calculator for a numerica approximation. If you don't care about the numeric approximation, you can use base 3 logs so that your denominator becomes 1.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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