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Solve  e^(x+2)ln(1-2x)=0

solve  e^(x+2)ln(1-2x)=0

Posted in Math, asked by feb, 6 years ago. 2160 hits.

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first, separate the power as follows:

xn(1-2x)+2n(1-2x)

then the equation becomes;

e^(xln(1-2x)+2ln(1-2x))=0

using the law of logarithms, we get

e^((n(1-2x)^x+ln(1-2x)^2)=0

using the law of indices to separate the LHS;

e^ln(1-2x)^x * e^ln(1-2x)^2 =0

now remember;

a*b=0→a=0 and b=0

also,e^lnx=x

(1-2x)^x * (1-2x)^2 =0

so,

(1-2x)^x=0 ..........1 and (1-2x)^2=0..................2

solving 1, remember 0^n=0

(1-2x)^x=0^x↔1-2x=0

hence, x=1/2

solving 2,

1-2x=±0

x=1/2 (repeated root)

PROOF

substitute x with 1/2 in the original equation, we get;

e^2.5ln(1-1)=e^ln(o)^2.5=e^ln(0)=0

David Owino
David Owino - 6 years ago
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