Question #39b7b

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Answer 1 · 2017-12-15T13:44:56

No solution.

#7^lnx = e^(lambda lnx)=(e^lnx)^lambda = x^lambda#

but #lambda = ln7# so finally we have

#x^ln7-x^ln7=0 ne 686# and the equations has not solution.

Answer 2 · 2017-12-15T13:47:04

There is no solution. Had the sign between the two terms been #+# instead of #-#, then there would be a solution. Check the problem.

We have #\ln(a^b)=b\ln a#, so:

#\ln(7^{\ln x})=(\ln x)(\ln 7)#

#\ln(x^{\ln 7})=(\ln 7)(\ln x)#

The logarithms are equal, therefore the numbers #7^{\ln x}# and #x^{\ln 7}# are equal, and their difference is never anything other than zero.