ln(x-2)+ln(2x-3)=2lnx solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use: e^ln(x-2) + e^ln(2x-3) = e^lnx^2 (x-2) + (2x-3) = x^^2 and then solve quadratic
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Originally Posted by attwoodp ln(x-2)+ln(2x-3)=2lnx solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use: e^ln(x-2) + e^ln(2x-3) = e^lnx^2 (x-2) + (2x-3) = x^^2 and then solve quadratic e^ln(x-2) + e^ln(2x-3) = e^lnx^2 (x-2) + (2x-3) = x^^2 it should be e^ln(x-2) * e^ln(2x-3) = e^lnx^2 (x-2)* (2x-3) = x^^2
Originally Posted by attwoodp ln(x-2)+ln(2x-3)=2lnx solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use: e^ln(x-2) + e^ln(2x-3) = e^lnx^2 (x-2) + (2x-3) = x^^2 and then solve quadratic You can't do that; what you meant to do is raise both sides to be the exponent of e, and then: $\displaystyle e^{ln(x-2)+ln(2x-3)} = e^{2lnx} \Rightarrow e^{ln( (x-2)(2x-3) )} = e^{ln (x^2)}$ This gives $\displaystyle (x-2)(2x-3) = x^2$ , and you know the rest...
Originally Posted by attwoodp ln(x-2)+ln(2x-3)=2lnx solve for x. Use log rules to convert the equation to: $\displaystyle \ln(2x^2\, -\, 7x\, +\, 6)\, =\, \ln(x^2)$ Then equate the arguments, and solve the resulting quadratic equation.
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