If log _{10}(x^{2}-6x+45)=2,then the values o | vedclass

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(D) Given the equation: $\log _{10}(x^{2}-6x+45)=2$By the definition of logarithm,$\log _{b}(a)=c$ implies $a=b^{c}$.Therefore,$x^{2}-6x+45=10^{2}$.$x

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$11, -5$

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(D) Given the equation: $\log _{10}(x^{2}-6x+45)=2$By the definition of logarithm,$\log _{b}(a)=c$ implies $a=b^{c}$.Therefore,$x^{2}-6x+45=10^{2}$.$x^{2}-6x+45=100$.Subtracting $100$ from both sides,we get $x^{2}-6x-55=0$.Factoring the quadratic equation: $(x-11)(x+5)=0$.Setting each factor to zero,we get $x-11=0$ or $x+5=0$.Thus,the values of $x$ are $x=11$ or $x=-5$.

If $\log _{10}(x^{2}-6x+45)=2$,then the values of $x$ are

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