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Question Number 147945 by Gbenga last updated on 24/Jul/21

25^x +5x=e^5 (can this be solved)

25x+5x=e5(canthisbesolved)

Commented by mr W last updated on 24/Jul/21

x=(e^5 /5)−((W(2×5^(((2e^5 )/5)−1) ln 5))/(2ln 5))≈1.536821

x=e55−W(2×52e55−1ln5)2ln5≈1.536821

Commented by Gbenga last updated on 24/Jul/21

thanks

thanks

Answered by Canebulok last updated on 24/Jul/21

Solution: ⇒ 25^x = e^5 −5x ⇒ (e^5 −5x)∙5^(−2x) = 1 ⇒ (e^5 −5x)e^(−2x∙ln(5)) = 1 By multiplying both sides by “ ((2∙ln(5))/5) ”, ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5))e^(−2x∙ln(5)) = ((2∙ln(5))/5) ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5))e^(((2e^5 ∙ln(5))/5)−2x∙ln(5)) = ((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) By Lambert W.function, ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5)) = W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ) ⇒ 2x∙ln(5) = ((2e^5 ∙ln(5))/5)−W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ) ⇒ x = ((((2e^5 ∙ln(5))/5)−W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ))/(2∙ln(5))) ⇒ x = (e^5 /5)−((W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ))/(2∙ln(5))) ≈ 1.536821146 Solution by: Kevin

Solution:⇒25x=e5−5x⇒(e5−5x)⋅5−2x=1⇒(e5−5x)e−2x⋅ln(5)=1Bymultiplyingbothsidesby‘‘2⋅ln(5)5″,⇒(2e5⋅ln(5)5−2x⋅ln(5))e−2x⋅ln(5)=2⋅ln(5)5⇒(2e5⋅ln(5)5−2x⋅ln(5))e2e5⋅ln(5)5−2x⋅ln(5)=2⋅ln(5)5⋅e2e5⋅ln(5)5ByLambertW.function,⇒(2e5⋅ln(5)5−2x⋅ln(5))=W(2⋅ln(5)5⋅e2e5⋅ln(5)5)⇒2x⋅ln(5)=2e5⋅ln(5)5−W(2⋅ln(5)5⋅e2e5⋅ln(5)5)⇒x=2e5⋅ln(5)5−W(2⋅ln(5)5⋅e2e5⋅ln(5)5)2⋅ln(5)⇒x=e55−W(2⋅ln(5)5⋅e2e5⋅ln(5)5)2⋅ln(5)≈1.536821146Solutionby:Kevin

Commented by Gbenga last updated on 24/Jul/21

nice solution

nicesolution

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