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How would solve the equation `3^(2x-1)` =`sqrt(3^x)?`
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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 by justaguide on October 1, 2013 at 2:56 PM (Answer #1)
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.
Divide each term in the equation by `3ln(3)` .
`x = 2/3`
Posted by kingattaskus12 on November 15, 2013 at 6:22 AM (Answer #2)
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