SOLUTION: This is a rather confusing question and I apologize if it confuses anyone else too.. but here goes:
e^(xlnb) (e^(xlnc)) = (bc)^2
Algebra.Com
Question 636327: This is a rather confusing question and I apologize if it confuses anyone else too.. but here goes:
e^(xlnb) (e^(xlnc)) = (bc)^2
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\left(e^{x\ln\,c}\right)\ =\ \left(bc\right)^2)
You didn't bother to share what you wanted so I can only presume that you want to solve the equation for 
. Actually, the only source of confusion is your failure to tell the whole story. Be that as it may:
Apply 
to the RHS:
}\ =\ \left(bc\right)^2)
Take the natural log of both sides:
}\right)\ =\ \ln\left(bc\right)^2)
Apply \ =\ n\,\cdot\,\log_b(a))
}\ln(e)\ =\ 2\ln\left(bc\right))
Apply \ =\ 1)
}\ =\ 2\ln\left(bc\right))
Apply the sum of the logs is the log of the product, \ +\ \log_b(a_2)\ =\ \log_b(a_1a_2))
}\ =\ 2\ln\left(bc\right))
Divide both sides by )
John
My calculator said it, I believe it, that settles it
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