`0.5^x-0.25=4` Solve the equation.

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For the given equation `0.5^x-0.25=4` , we may simplify by combining like terms.

Add `0.25` on both sides of the equation.

`0.5^x-0.25+0. 25=4+0.25`

`0.5^x=4.25`

Take the "`ln` " on both sides to be able to bring down the exponent value.

Apply the natural logarithm property: `ln(x^n)= n*ln(x)` .

`ln(0.5^x)=ln(4.25)`

`xln(0.5)=ln(4.25)`

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For the given equation `0.5^x-0.25=4` , we may simplify by combining like terms.

Add `0.25` on both sides of the equation.

`0.5^x-0.25+0. 25=4+0.25`

`0.5^x=4.25`

Take the "`ln` " on both sides to be able to bring down the exponent value.

Apply the natural logarithm property: `ln(x^n)= n*ln(x)` .

`ln(0.5^x)=ln(4.25)`

`xln(0.5)=ln(4.25)`

To isolate the x, divide both sides by `ln(0.5)` .

`(xln(0.5))/(ln(0.5))=(ln(4.25))/(ln(0.5))`

`x=(ln(4.25))/(ln(0.5))`

`x=(ln(17/4))/(ln(1/2))`

`x=(ln(17) -ln(4))/(ln(2^(-1)))`

`x=(ln(17) -ln(2^2))/(ln(2^(-1)))`

`x=(ln(17) -2ln(2))/(-ln(2))`

`x=(ln(17))/(-ln(2)) -(2ln(2))/(-ln(2))`

`x= -(ln(17))/(ln(2)) +2 or -2.087` (approximated value)

Checking: Plug-in `x=-2.087` on `0.5^x-0.25=4` .

`0.5^(-2.087)-0.25=?4`

`(1/2)^(-2.087)-0.25=?4`

`(2^(-1))^(-2.087)-0.25=?4`

`2^((-1)*(-2.087))-0.25=?4`

`2^(2.087)-0.25=?4`

`4.25-0.25=?4`

`4=4 `   TRUE

Note: `2^(2.087)=4.248636746 ~~4.25`

Therefore,there is no extraneous solution.

The `x=-(ln(17))/(ln(2)) +2`   is the real exact solution of the given equation `0.5^x-0.25=4` .

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