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# How would solve the equation `3^(2x-1)` =`sqrt(3^x)?`

ruals | Salutatorian

Posted October 1, 2013 at 2:51 PM via web

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How would solve the equation `3^(2x-1)` =`sqrt(3^x)?`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 1, 2013 at 2:56 PM (Answer #1)

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The equation `3^(2x - 1) = sqrt(3^x)` has to be solved.

`3^(2x - 1) = sqrt(3^x)`

=> `3^(2x - 1) = 3^(x/2)`

As the base is the same equate the exponent

`2x - 1 = x/2`

=> `(3x)/2 = 1`

=> x = `2/3`

The solution of the equation is `x = 2/3`

Posted November 15, 2013 at 6:22 AM (Answer #2)

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To solve `3^(2x - 1) = sqrt(3^x)` . Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

`ln(3^(2x - 1)) = ln (sqrt(3^x))`

Remove the parentheses that are not needed from the expression.

`2x ln(3) - ln(3) = 1/2 x ln(3)`

Move all terms not containing `x` to the right-hand side of the equation.

`-ln (3) + (3 x ln (3))/2 = 0`

Since `-ln(3)` does not contain the variable to solve for, move it to the right-hand side of the equation by adding `ln(3)` to both sides.

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

Multiply `ln(3)` by each term inside the parentheses.

`3xln(3)=2ln(3) `

Divide each term in the equation by `3ln(3)` .

Therefore,

`x = 2/3`

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