Question 1041843: please help solve this x : 2*7^(x-5)=3E^(x+2)
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Your equation is nonsense unless I assume that E is supposed to represent  , the base of the natural logarithms. Never, ever, use E to represent .
Take the log of both sides. Whether you use base 10 logs or natural logs or any other base that tickles your fancy won't matter in the final analysis since the numerical answer you get will be the same. However, since you have raised to a power in the equation it will be more convenient to take the natural log.
\ =\ \ln\left(3e^{x\,+\,2}\right))
Apply the rules of logarithms: The sum of the logs is the log of the product, the difference of the logs is the log of the quotient, the log of a base to a power is the power times the log of the base, and the log of a base is equal to 1.
\ + \ln\left(7^{x\,-\,5}\right)\ =\ \ln\left(3\right)\ +\ \ln\left(e^{x\,+\,2}\right))
\ + (x\,-\,5)\ln\left(7\right)\ =\ \ln\left(3\right)\ +\ (x\,+\,2)\ln\left(e\right))
\ + (x\,-\,5)\ln\left(7\right)\ =\ \ln\left(3\right)\ +\ (x\,+\,2))
The rest is elementary algebra; isolate the variable by adding the additive inverses and multiplying by the reciprocal. Also using logarithm rules in reverse. Remember, the log of a constant is just another number.
\ln\left(7\right)\ -\ (x\,+\,2)\ =\ \ln\left(3\right)\ -\ \ln\left(2\right)\ +\ (x\,+\,2))
\ -\ \ln\left(7^5\right)\ -\ x\ -\ 2\ =\ \ln(3)\ -\ \ln(2))
\ -\ 1\right)\ =\ \ln(3)\ +\ \ln\left(7^5\right)\ -\ \ln(2)\ +\ 2)
\ -\ 1\right)\ =\ \ln\left(\frac{3\,\cdot\,7^5}{2}\right)\ +\ 2)
\ +\ 2}{\ln(7)\ -\ 1}=\ \frac{\ln\left(25210.5\right)\ +\ 2}{\ln(7)\ -\ 1})
Is the exact answer. Use your calculator if you need a numeric approximation.
John
My calculator said it, I believe it, that settles it
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