Logarithms - Get the second solution to a logarithm equationHere's the original equation:ln⁡(11x−10)+(ln⁡(11x−10))2=6I've managed to obtain one solution: x=e2+1011through those steps:1. ln⁡((11x−10)(11x−10)2)=62. ln⁡((11x−10)3)=63. (11x−10)3=e64. 11x−10=e25. 11x=e2+106. x=e2+1011The textbook shows a second solution: x=10e3+111e3, but how do you get to that result?

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Logarithms - Get the second solution to a logarithm equationHere&#039;s the original equation:ln⁡(11x−10)+(ln⁡(11x−10))2=6I&#039;ve managed to obtain one solution: x=e2+1011through those steps:1. ln⁡((11x−10)(11x−10)2)=62. ln⁡((11x−10)3)=63. (11x−10)3=e64. 11x−10=e25. 11x=e2+106. x=e2+1011The textbook shows a second solution: x=10e3+111e3, but how do you get to that result?

Demetrius Kaufman · Accepted Answer

Your solution is incorrect because :ln⁡(11x−10)+(ln⁡(11x−10))2 is not equal toln⁡(11x−10×(11x−10)2)To solveln⁡(11x−10)+(ln⁡(11x−10))2=6Let t=ln⁡(11x−10) t2+t−6=0this gives , t=2 and t=−3when t=2→ln⁡(11x−10)=2→x=e2+1011when t=−3→ln⁡(11x−10)=−3→x=10e3+111e3

Ronald Martinez · Accepted Answer

Hint: Let a=ln⁡(11x−10). Then you would havea+a2=6ora2+a−6=0.

Logarithms - Get the second solution to a

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