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Hi! could you please solve for x this equation? `2^(log 2x)=3^(log3x)`  thank you

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blehens | (Level 1) eNoter

Posted September 4, 2013 at 1:46 AM via web

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Hi! could you please solve for x this equation?

`2^(log 2x)=3^(log3x)` 

thank you

2 Answers | Add Yours

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 4, 2013 at 6:21 AM (Answer #2)

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The equation `2^(log 2x) = 3^(log 3x)` has to be solved for x.

`2^(log 2x) = 3^(log 3x)`

Take the logarithm of both the sides

`log(2^(log 2x)) = log(3^(log 3x))`

Use the property `log a^b = b*log a`

=> `log 2x*log 2 = log 3x*log 3`

Use the property `log a*b = log + log b`

=> `log 2*(log x + log 2) = log 3((log 3 + log x)`

=> `log x(log 2 - log 3) = (log 3)^2 - (log 2)^2`

=> `log x*log(2/3) = (log 3 + log 2)(log 3 - log 2)`

=> `log x = (log 6*log 1.5)/(log 2/3)`

=> `log x ~~ -0.7781`

x = `10^-0.7781 = 1/6`

The solution of the equation is `x = 1/6`

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aruv | High School Teacher | (Level 2) Valedictorian

Posted September 4, 2013 at 4:36 AM (Answer #1)

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`2^(log(2x))=3^(log(3x))`

Taking natural log both side

`ln(2^(log(2x)))=ln(3^(log(3x)))`

`log(2x)ln(2)=log(3x)ln(3)`

`.6931log(2x)=1.0986log(3x)`

`.6931(log(2)+log(x))=1.0986(log(3)+log(x))`

`.6931(.3010+log(x))=1.0986(.4771+log(x))`

`.2086+.6931log(x)=.5242+1.0986log(x)`

`.4055 xx log(x)=.3156`

`log(x)=.3156/.4055`

`log(x)=.7783`

Taking antilog both side

`x=6.002`

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