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How do I solve `4 ^(-x+4) = 31` ?

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monique06 | (Level 3) Valedictorian

Posted June 5, 2013 at 1:42 PM via web

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How do I solve `4 ^(-x+4) = 31` ?

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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted June 5, 2013 at 2:01 PM (Answer #1)

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`4^(-x+4)=31`

Since the variable x is part of the exponent, to solve, it has to be removed from the exponent.

To do so, take the logarithm of both sides. Note that the logarithm can have any base. For this problem, let's use the natural logarithm.

`ln 4^(-x+4)=ln31`

At the left side, apply the property `ln a^m = m ln a` .

`(-x+4)ln4=ln31`

Now that the x is no longer part of the exponent, let's isolate it. So, divide both sides by ln4.

`((-x+4)ln4)/ln4=(ln31)/(ln4)`

`-x+4=(ln31)/(ln4)`

And, subtract both sides by 4.

`-x+4-4=(ln31)/(ln4)-4`

`-x=(ln31)/(ln4)-4`

To have a positive x at the left side, divide both sides by -1.

`(-x)/(-1)= ((ln31)/(ln4)-4)/(-1)`

`x=-((ln31)/(ln4)-4)`

` `

Hence, the solution to the equation is` x=1.5229` .

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