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Solve The Equations Below, Round to the nearest tenth-thousandth.
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\[\log_{10}x=\log_{10}2x ^{2} -2\]
@ganeshie8
@hartnn
Please Show Your Steps.
Because the bases are the same (both are equal to 10), we can just set the insides of the log equal to each other \[x = 2x^2 -2\] Solve for x \[2x^2 -x -2 = 0\] Now you can factor or use the quadratic equation to solve for x.
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(2x-2) (x- 1) ?
Use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] From the equation I wrote above \[2x^2 - x - 2 = 0\] \[a =2, \ b = -1, \ c = -2\] So just plug those values into the quadratic equation and solve. Keep in mind the instructions, which ask you to round to the nearest ten thousandth.
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