If log _{10}left[98+sqrt{x^{2}-12 x+36}right] | vedclass

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(D) Given the equation: $\log _{10}\left[98+\sqrt{x^{2}-12 x+36}\right]=2$By the definition of logarithm,if $\log_{b}(a) = c$,then $a = b^c$. Applying

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$4, 8$

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(D) Given the equation: $\log _{10}\left[98+\sqrt{x^{2}-12 x+36}\right]=2$By the definition of logarithm,if $\log_{b}(a) = c$,then $a = b^c$. Applying this:$98+\sqrt{x^{2}-12 x+36} = 10^2$$98+\sqrt{(x-6)^2} = 100$$\sqrt{(x-6)^2} = 100 - 98$$|x-6| = 2$This gives two cases:Case $1$: $x-6 = 2 \Rightarrow x = 8$Case $2$: $x-6 = -2 \Rightarrow x = 4$Thus,the values of $x$ are $4$ and $8$.

If $\log _{10}\left[98+\sqrt{x^{2}-12 x+36}\right]=2,$ then $x=$

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